CBSE Chemistry Solutions Class 12 Important Questions 2024-25

Very Short Questions and Answers (1 Mark Questions)

1. Give an example of ‘liquid in solid’ type solution.

Ans: Liquid in solid means solvent will be solid and the solute will be liquid. Example, mercury in sodium (amalgam).

2. Which type of solid solution will result by mixing two solid components with  large differences in the sizes of their molecules?

Ans: A solid solution which will result by mixing two solid components with large difference in the sizes of their molecules is known as interstitial solid solution.

3. What is meant by semimolar and decimolar solutions?

Ans: The semimolar and decimolar terms are completely based on the molarity of the solution. Semimolar signifies the molarity of solution as $\frac{M}{2}$ and decimolar signifies the molarity of solution as $\frac{M}{10}$ .

4. What will be the mole fraction of  water in ${{C}_{2}}{{H}_{5}}OH$  solution containing equal number of moles of water and ${{C}_{2}}{{H}_{5}}OH$?

Ans: Mole fraction of any component is given as the ratio of the respective moles of the same to the total moles of the solution. Here, if we consider x moles of water then we will have equal i.e., x moles of ${{C}_{2}}{{H}_{5}}OH$ . Thus,

\[\text{Mole fraction = }\frac{x}{x+x}=\frac{1}{2}=0.5\] 

5. Which of the following is a dimensionless quantity: molarity, molality or  mole fraction?

Ans: Out of molarity, molality and mole fraction; mole fraction is the dimensionless quantity as it is the ratio of similar quantities i.e., moles.

6. 10 g glucose is dissolved in 400 g of solution. Calculate percentage concentration of the solution.

Ans: In general, the percentage is the ratio of a component to the total solution multiplied by 100. Thus, the percentage concentration of the solution is given as,

\[\text{Percentage of concentration of solution = }\frac{\text{10}}{\text{400}}\text{ }\!\!\times\!\!\text{ 100=2}\text{.5%}\]

7. Gases tend to be less soluble in liquids as the temperature is raised. Why?

Ans: When gases are dissolved in liquid, they have high velocities due to their free movement; rising the temperature shifts the equilibrium of the reaction in the backward direction which then makes gases less soluble in liquid.

8. State the conditions which must be satisfied if an ideal solution is to be formed.

Ans: If an ideal solution is to be formed, the primary condition to be fulfilled will be a solution obeying Raoult’s law. Also, the solution should be homogenous and there must not be any volumetric or thermal effects.

9. A mixture of chlorobenzene and bromobenzene forms a nearly ideal solution  but a mixture of chloroform and acetone does not. Why?

Ans: In the case of a mixture of chlorobenzene and bromobenzene, there is equal interaction of the molecules hence, forming an ideal solution. Whereas, they are absent in the mixture of chloroform and acetone.  

10. How is the concentration of a solute present in trace amount in a solution  expressed?

Ans: When the solution has trace amounts of solute present in them, the concentration of the solute is expressed in parts per million i.e., ppm.

11. ${{N}_{2}}$  and ${{O}_{2}}$  gases have ${{K}_{H}}$  values 76.48 kbar and 34.86 kbar respectively at 293 K temperature. Which one of these will have more solubility in water?

Ans: In general, less the value of ${{K}_{H}}$ greater the solubility of a gas at the same partial pressure. Here, we can see ${{O}_{2}}$ has less value of ${{K}_{H}}$therefore, it will be more soluble than ${{N}_{2}}$ .

12. Under what condition molality and molarity of a solution are identical. Explain with suitable reasons.

Ans: The condition where the molarity and molality of the solution will be identical, is the presence of high amounts of solvent in the same i.e., dilute solutions.

13. Addition of $\mathbf{Hg{{I}_{2}}}$  to KI (aq.) shows increase in vapor pressure. Why?

Ans: Vapor pressure is the colligative property which is completely dependent on the number of particles the solution has. When $Hg{{I}_{2}}$ is added to KI, they form a complex decreasing the number of solute particles. Thus, the vapor pressure will increase.

14. What will happen to the boiling point of the solution formed on mixing two miscible liquids showing negative deviation from Raoult’s law?

Ans: When the two miscible liquids showing negative deviation from Raoult’s law are mixed forming a solution, then the vapor pressure will decrease increasing its boiling point.

15. Liquid ‘Y’ has higher vapor pressure than liquid ‘X’, which of them will have higher boiling point?

Ans: In general, high vapor pressure corresponds to the low boiling point of the liquid. Here, the vapor pressure of liquid ‘X’ is the lowest thus, it will have the highest boiling point.

16. When 50 mL of ethanol and 50 mL of water are mixed, predict whether the volume of the solution is equal to, greater than or less than 100 mL.  Justify.

Ans: When the equal volume of ethanol and water are mixed together i.e., 50 ml each then the total volume formed will be less than 100 ml. This is because the molecules of ethanol are smaller than that of water which makes it fit themselves within the big water molecules. 

17. Which type of deviation is shown by the solution formed by mixing cyclohexane and ethanol?

Ans: The solution formed by mixing cyclohexane and ethanol will show positive deviation from Raoult’s law.

18. A and B liquids on mixing produce a warm solution. Which type of deviation from Raoult’s law is there?

Ans: When mixing A and B produces a warm solution, then the solution will show negative deviation from Raoult’s law.

19. Define cryoscopic constant (molal freezing point depression constant.)

Ans: The freezing point depression on dissolving a non – volatile solute in 1 Kg solvent is known as cryoscopic constant or molal freezing point depression constant. 

20. Mention the unit of ebullioscopic constant (molal boiling point elevation constant.)

Ans: The unit of ebullioscopic constant or molal boiling point elevation constant is $\text{K kg mo}{{\text{l}}^{-1}}$.

21. If ${{K}_{f}}$  for water is $1.86 gmo{{l}^{-1}}$ , what is the freezing point of 0.1 molal solution of a substance which undergoes no dissociation or association of solute? 

Ans: The freezing point depression is given as,

\[\Delta {{T}_{f}}=i{{K}_{f}}m\] 

Here, i = 0 as there is no dissociation or association of solute. Thus,

\[\Delta {{T}_{f}}={{K}_{f}}m\] 

\[\Delta {{T}_{f}}=1.86\times 0.1=0.186K\] 

22. What is reverse osmosis? Give one large scale use of it.

Ans: The process taking place opposite to that of the osmosis (solvent flow towards solution through semi – permeable membrane) by applying pressure greater than osmotic pressure on the solution. The most common large scale application is desalination of sea water.

23. What is the maximum value of van’t Hoff factor (i) for $\mathbf{N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O}$? 

Ans: The maximum value of Van’t Hoff factor (i) for $N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$ will be 3.

24. What is the value of van’t Hoff factor (i) if solute molecules undergo dimerisation.

Ans: The value of Van’t Hoff factor (i) for a solute who undergoes dimerization will be 0.5.

25. Under what condition is van’t Hoff factor less than one?

Ans: Under association of molecules only, the Van’t Hoff factor will be less than 1.

26. The Phase Diagram for pure solvent and the solution containing non - volatile  solute are recorded below. The quantity indicated by ‘X’ in the figure is known as:

(image will be uploaded soon)

Ans: The quantity ‘X’ as indicated in the figure indicates $\Delta {{T}_{b}}$ .

27. $\mathbf{AgN{{O}_{3}}}$  on reaction with NaCl in aqueous solution gives white precipitate. If the two solutions are separated by a semi-permeable membrane, will there be appearance of a white ppt. in the side ‘X’ due to osmosis? 

(image will  be uploaded soon)

Ans: If the solutions of $AgN{{O}_{3}}$ and NaCl are separated by a semipermeable membrane there will be no appearance of ppt due to the movement of only solvent particles through the membrane and not of solute particles. 

Short Questions and Answers (2 Marks Questions)

1. Explain the following:

1. Solubility of a solid in a liquid involves dynamic equilibrium.

Ans: The solubility of solid in the liquid is maximum concentration of solute to be dissolved in solvent. Any extra solute will lead to the precipitation of the dissolved solute which leads to the attainment of dynamic equilibrium.

2. Ionic compounds are soluble in water but are insoluble in nonpolar  solvents.

Ans: In general, we know the empirical rule of ‘like dissolves like’. Here, the ionic compounds are polar molecules which are soluble in water due to its polar nature.  

2. Give two examples each of a solution:

1. Showing positive deviation from Raoult’s Law.

Ans: The mixture of 

(i) Ethanol and water.

(ii) Acetone and ethanol. 

2. Showing negative deviation from Raoult’s Law.

Ans: The mixture of

(i) HCl and water.

(ii) Acetone and chloroform.

3. Draw vapor pressure vs composition (in terms of mole fraction) diagram for an ideal solution.

Ans:

4. Define azeotropes with one example of each type.

Ans: The binary mixtures of solutions that have the same composition in liquid and vapor phase having constant boiling points throughout the distillation are known as azeotropes. 

1. Minimum Boiling Azeotropes – Mixture of water and ethanol (solutions showing positive deviation from Raoult’s law).

2. Maximum Boiling Azeotropes – Mixture of water and HCl (solutions showing negative deviation from Raoult’s law).

5. Draw the total vapor pressure vs. mol fraction diagram for a binary solution exhibiting non-ideal behavior with negative deviation.

Ans:

6. The vapor pressure curve for three solutions having the same non- volatile solute in the same solvent are shown. The curves are parallel to each other and do not intersect. What is the correct order of the concentrations of the solutions? (Hint: A < B < C)

(image will be uploaded soon)

Ans: As shown in the figure, the concentration of the three solutions having the same non – volatile solute in the same solvent will be in the order of A < B < C.

7. Show that the relative lowering of vapor pressure of a solvent is a colligative  property.

Ans: Colligative properties depend upon the number of solute particles in the solution. To be more precise, it is the ratio of the number of solute particles to the number of solvent particles in the solution.

Now, relative lowering of vapour pressure is given as; 

\[\frac{\Delta P}{{{P}^{0}}}=\frac{{{P}^{0}}-P}{{{P}^{0}}}\] 

Also, by Raoult’s law we know that,

\[\frac{{{P}^{0}}-P}{{{P}^{0}}}=x\] 

where, x is the mole fraction of any component.

Hence, we can say that the relative lowering of vapor pressure of a solvent is a colligative  property.

8. Benzene and toluene form a nearly ideal solution. At a certain temperature, calculate the vapor pressure of solution containing equal moles of the two substances. [Given: P0Benzene  = 150 mm of Hg, $P_{Toluene}^{0}$  = 55 mm of Hg]

Ans: In accordance with the given data, we know that,

\[{{P}_{Solution}}=P_{Benzene}^{0}\left[ \frac{n}{n+n} \right]+P_{Toluene}^{0}\left[ \frac{n}{n+n} \right]\] 

As, we have given equal moles of both the substances. Thus,

\[{{P}_{Solution}}=\left( P_{Benzene}^{0}+P_{Toluene}^{0} \right)\times \frac{1}{2}\] 

\[{{P}_{Solution}}=\left( 150+55 \right)\times \frac{1}{2}=102.5mmHg\] 

9. What is meant by abnormal molecular mass? Illustrate it with suitable examples.

Ans: The molecular masses which are different theoretical molecular masses due to either association or dissociation of molecules in the solution are known as abnormal molecular masses. For example, weak electrolytes which dissociates to form ions. 

10. When 1 mole of NaCl is added to 1 liter water, the boiling point increases? When 1 mole of $C{{H}_{3}}OH$  is added to 1 liter water, the boiling point decreases? Suggest reasons.

Ans: When NaCl (non – volatile solute) is added to water then, the vapor pressure of the solution is lowered increasing the boiling point. Whereas, $C{{H}_{3}}OH$ is more volatile than water i.e., has more vapor pressure than that of water decreasing the boiling point. 

11. Can we separate water completely from $HN{{O}_{3}}$  solution by vaporization? Justify your answer.

Ans: Water cannot be separated from nitric acid by evaporation or vaporization due to the formation of azeotropic mixture which makes the separation impossible.

12. 1 gram each of two solutes ‘A’ and ‘B’ (molar mass of A > molar mass of B) are dissolved separately in 100 g each of the same solvent. Which solute will show greater elevation in boiling point and why?

Ans: The elevation in boiling point is given as,

\[\Delta {{T}_{b}}={{K}_{b}}m={{K}_{b}}\frac{n}{W}\] 

where,

n = number of moles of a component given as,

\[n=\frac{\text{mass of component}}{\text{molar mass of component}}\] 

W = weight of solvent in Kgs.

In accordance to the given data and the proportionality we can say that,

\[\Delta {{T}_{b}}\propto \frac{1}{\text{molar mass of component}}\] 

Thus, for A and B;

\[\Delta T_{b}^{A}=\frac{1}{{{M}_{A}}}\text{ and }\Delta T_{b}^{B}=\frac{1}{{{M}_{B}}}\] 

Dividing an equation of A by B we get,

\[\frac{\Delta T_{b}^{A}}{\Delta T_{b}^{B}}=\frac{{{M}_{B}}}{{{M}_{A}}}\] 

We have given that the molar mass of A > molar mass of B thus, B will show greater elevation in boiling point.

13. Examine the following illustrations and answer the following questions

(image will be uploaded soon)

1. Identify the liquid A and liquid B (pure water or sugar solution).

Ans: Liquid A is sugar solution and liquid B is pure water.

2. Name the phenomenon involved in this experiment so that the level of liquid in this the funnel has risen after some time.

Ans: The ongoing phenomenon is the osmosis process in which the level of liquid in the funnel has risen after some time. 

14. How relative lowering in vapor pressure is related with depression in freezing point and elevation in boiling point?

Ans: The relative lowering in vapor pressure, depression in freezing point and elevation in boiling point; all the three are the colligative properties of the solutions and inter – related. The colligative properties completely depend upon the solute particles dissolved in solvent. So, when the non – volatile solute is added to the solvent due to change in vapor pressure, all other properties change. The relative lowering in vapor pressure of solute brings depression in freezing point and elevation in boiling point of the same. 

Long Questions and Answers (3 Marks Questions)

1. (a)   State Henry’s Law.

Ans: A gas law which states that the amount of gas dissolved in liquid is directly proportional to the partial pressure of the same gas above the liquid level when the temperature is kept constant is known as Henry’s law. Mathematically, it is stated as;

\[P={{K}_{H}}\times C\] 

where,

C = concentration of dissolved gas in the liquid

\[{{K}_{H}}~=\text{ }Henrys\text{ }law\text{ }constant\] 

P = partial pressure of the gas 

(b)  If ${{O}_{2}}$  is bubbled through water at 393 K, how many millimoles of ${{O}_{2}}$ gas would be dissolved in 1L of water? Assume that ${{O}_{2}}$ exerts a pressure of 0.95 bar. (Given ${{K}_{H}}$  for ${{O}_{2}}$ = 46.82 bar at 393K).

Ans: In accordance with the given data and formula for Henry’s law;

Concentration of gas,

\[C=\frac{P}{{{K}_{H}}}=\frac{0.95}{46.82}=0.0202M\] 

Now, number of moles is given as;

\[n=M\times V\] 

\[n=0.0202\times 1=0.0202moles\] 

Thus, moles of ${{O}_{2}}$ dissolved in water are 20.2 millimoles.

2. Given reason for the following: –

1. Aquatic species are more comfortable in cold waters than in warm  waters.

Ans: The solubility of oxygen decreases with increase in temperature which makes aquatic organisms feel comfortable in cold water due to high amounts of dissolved oxygen.

2. To avoid bends scuba divers use air diluted with helium.

Ans: The bends during scuba diving is a condition which is very painful and dangerous for life. Thus, divers use helium as a dissolved gas in the air due to its low solubility in human blood.

3. Cold drinks bottles are sealed under high pressure of $C{{O}_{2}}$ .

Ans: The $C{{O}_{2}}$ gas is filled in the cold drinks under high pressure and sealed as with high pressure, the solubility of gases in liquid increases.

3. Why should a solution of a non-volatile and non-electrolyte solute boil at  a higher temperature? Explain with the help of a diagram. Derive the  relationship between molar mass and elevation in boiling point.

Ans: The solution of a non – volatile and non – electrolyte solute should boil at high temperature as their addition lowers the vapor pressure causing them to boil only at higher temperature i.e., higher than the boiling point of pure solvent. This can be explained by the below given diagram;

(image will be uploaded soon)

The relationship between the molar mass of solute and the elevation in boiling point is given as;

Elevation in boiling point,

\[\Delta {{T}_{b}}={{K}_{b}}m\] 

where, m = molality and is stated as;

\[m=\frac{{{W}_{2}}}{{{M}_{2}}{{W}_{1}}}\] 

Now, putting the values in empirical equation we get,

\[{{M}_{2}}=\frac{{{K}_{b}}{{W}_{2}}}{\Delta {{T}_{b}}{{W}_{1}}}\] 

4. Account for the following: –

1. $CaC{{l}_{2}}$  is used to clear snow from roads in hill stations.

Ans: The calcium chloride is responsible for depression in freezing point of water to an extent from where it cannot form ice hence, clearing the snow from roads. 

2. Ethylene glycol is used as antifreeze solution in radiators of vehicles  in cold countries.

Ans: The addition of ethylene glycol in car radiators lowers the freezing point of water so as to avoid bursting of lines in cold winter temperatures.

3. The freezing point depression of 0.01 m NaCl is nearly twice that of 0.01 m glucose solution.

Ans: The sodium chloride dissociates in the solution giving two ions and glucose will never dissociate to form ions. Thus, by the relationship of freezing point depression;

\[\Delta {{T}_{f}}=i{{K}_{f}}m\] 

NaCl can have nearly twice the depression in freezing point to that of glucose even though the concentrations are the same.

5. Why do colligative properties of solution of a given concentration are found to give abnormal molecular weight of solute. Explain with the help of suitable examples.

Ans: The colligative properties have been found to give abnormal molecular masses due to the presence of fact about association or dissociation of solute molecules in solvent. To help this out Van’t Hoff factor was introduced which properly explained the fact. 

Solute undergoing association in the solution, Van’t Hoff factor is less than 1. Whereas, solute undergoing dissociation in the solution has Van’t Hoff factor more than 1.

Example:

\[KCl\leftrightarrow {{K}^{+}}+C{{l}^{-}}\] 

Here, a molecule of KCl gives 2 ions which gives Van’t Hoff factor (i) as 2. This will reflect in the form of abnormal molecular mass calculations for certain conditions. 

6. Give reasons for the following: –

1. RBC swells up and finally bursts when placed in 0.1% NaCl solution.

Ans: When the RBCs are placed in the 0.1% NaCl solution the water will flow inside the cells as it moves from less concentrated to more concentrated solution. At a point the cell will be so swollen, enough to burst.  

2. When fruits and vegetables that have been dried are placed in water, they slowly swell and return to original form.

Ans: The drying of fruits or vegetables causes them to become more concentrated. When these are placed in water, the phenomenon of osmosis takes place resulting in the original form of those fruits and vegetables.

3. A person suffering from high blood pressure is advised to take less amount of table salt.

Ans: The intake of NaCl can result in increasing the number of solutes in the body fluids. If the solute concentration is increased, the osmotic pressure increment can rupture the blood cells. 

7. Glycerin, ethylene glycol and methanol are sold at the same price per kg. Which would be cheaper for preparing an antifreeze solution for the radiator of an automobile?

Ans: The antifreeze solution is responsible for lowering the freezing point of water so as to avoid the damage in the radiator in automobile 

system. Here, 3 solutes are needed to be compared i.e., glycerin, 

ethylene glycol and methanol. 

According the formula;

\[\Delta {{T}_{f}}={{K}_{f}}m\] 

where, m = molality and is stated as;

\[m=\frac{{{W}_{2}}}{{{M}_{2}}{{W}_{1}}}\] 

We can say that the lower molar mass will give depression in freezing point and hence, the best results at lower amounts.

Molar masses are given as;

Glycerin = 92 g/mol

Ethylene glycol = 62 g/mol

Methanol = 32 g/mol

Thus, methanol will be cheaper for preparing an antifreeze solution for the radiator of an automobile.

8. Determine the correct order of the property mentioned against them:

1. 10% glucose (${{p}_{1}}$ ), 10% urea (${{p}_{2}}$ ), 10% sucrose (${{p}_{3}}$ ) [Osmotic pressure]

Ans: 10% urea > 10% glucose > 10% sucrose.

As, the osmotic pressure depends upon the molar concentration of the solute; urea has the highest molar concentration i.e., more moles in the solution than that of glucose and sucrose.

2. 0.1 m NaCl, 0.1 m urea, 0.1 m $MgC{{l}_{2}}$ (Elevation in b.pt.)

Ans: 0.1 m $MgC{{l}_{2}}$ > 0.1 m NaCl > 0.1 urea.

As, the elevation in boiling point depends upon the Van’t Hoff factor; $MgC{{l}_{2}}$ will dissociate producing highest number of ions (3) than that of NaCl (2) and urea(0).

3. 0.1 m $CaC{{l}_{2}}$ , 0.1 m sucrose, 0.1 m NaCl (Depression in f.pt.)

Ans: 0.1 m $CaC{{l}_{2}}$ > 0.1 m NaCl > 0.1 m sucrose.

As, the depression in freezing point depends upon the Van’t Hoff factor; $CaC{{l}_{2}}$ will dissociate producing the highest number of ions (3) than that of NaCl (2) and sucrose(0).

9. For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is $2{}^\circ C$ . Assuming concentration of solute is much lower than the concentration of solvent, determine the vapor pressure (mm of Hg) of the solution. 

(Given: ${{K}_{b}}$  for water = $0.76Kkgmo{{l}^{-1}}$ )

Ans: We have given that,

\[{{W}_{2}}=2.5g\] 

\[{{W}_{1}}=100g\] 

\[\Delta {{T}_{b}}=2{}^\circ C\] 

\[{{P}^{0}}=1atm=760mmHg\] 

\[{{K}_{b}}=0.76Kkgmo{{l}^{-1}}\] 

\[{{M}_{1}}=18gmo{{l}^{-1}}\] 

Now, we have elevation in boiling point as;

\[\Delta {{T}_{b}}=\frac{{{K}_{b}}{{W}_{2}}}{{{M}_{2}}{{W}_{1}}}\] 

\[\Delta {{T}_{b}}=\frac{0.76\times 1000\times 2.5}{{{M}_{2}}\times 100}=2\] 

\[{{M}_{2}}=9.5gmo{{l}^{-1}}\] 

Now, for dilute solutions; \[\frac{{{P}^{0}}-{{P}_{S}}}{{{P}^{0}}}=\frac{{{W}_{2}}{{M}_{1}}}{{{M}_{2}}{{W}_{1}}}\] 

\[\frac{760-{{P}_{S}}}{760}=\frac{2.5\times 18}{9.5\times 100}=0.047\] 

Thus, the vapor pressure of the solution, ${{P}_{S}}=724mmHg$ 

10. 15.0 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution was fund to freeze at $-0.34{}^\circ C$ . What is the molar mass of this substance? (${{K}_{f}}$  for water = $1.86Kkgmo{{l}^{-1}}$ ). 

Ans: We have given that,

\[{{W}_{2}}=15g\] 

\[{{W}_{1}}=450g\] 

\[\Delta {{T}_{f}}=0.34{}^\circ C\] 

\[{{K}_{f}}=1.86Kkgmo{{l}^{-1}}\] 

Now, we have elevation in boiling point as;

\[\Delta {{T}_{f}}=\frac{{{K}_{f}}{{W}_{2}}}{{{M}_{2}}{{W}_{1}}}\] 

\[\Delta {{T}_{f}}=\frac{1.86\times 1000\times 15}{{{M}_{2}}\times 450}=0.34\] 

\[{{M}_{2}}=182.35gmo{{l}^{-1}}\] 

Very Long Questions and Answers (5 Marks Questions)

1. (a) What are ideal solutions? Write two examples.

Ans: The solutions obeying Raoult’s law at any concentration and temperature are known as ideal solutions. For example;

(i) Benzene and toluene

(ii) Ethyl bromide and ethyl iodide.

(b) Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0g of polymer of molar mass 185000 in 450 mL of water at $37{}^\circ C$ .

Ans: We have given that,

\[W=1.0g\] 

\[M=185000gmo{{l}^{-1}}\] 

\[V=450ml=0.45L\] 

\[T=37+273=310K\] 

\[R=8.314Jmo{{l}^{-1}}{{K}^{-1}}=8314PaLmo{{l}^{-1}}{{K}^{-1}}\] 

Osmotic pressure is given as;

\[\pi =\frac{WRT}{MV}\] 

\[\pi =\frac{8314\times 310}{185000\times 0.45}=30.95Pa\] 

2. (a) Describe a method of determining molar mass of a non - volatile solute from vapor pressure lowering.

Ans: There is a very general method of determining molar mass of a    non – volatile solute from vapor pressure. 

By Raoult’s law;

\[\frac{{{P}^{0}}-{{P}_{S}}}{{{P}^{0}}}=x\] 

\[\frac{{{P}^{0}}-{{P}_{S}}}{{{P}^{0}}}=\frac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}\] 

If ${{n}_{2}}<<{{n}_{1}}$ then,

\[\frac{{{P}^{0}}-{{P}_{S}}}{{{P}^{0}}}=\frac{{{n}_{2}}}{{{n}_{1}}}\] 

Thus,

\[\frac{{{P}^{0}}-{{P}_{S}}}{{{P}^{0}}}=\frac{{{W}_{2}}{{M}_{1}}}{{{M}_{2}}{{W}_{1}}}\] 

From this we can easily determine the ${{M}_{2}}$ .

(b) How much urea (mol. mass 60 g/mol) must be dissolved in 50g of water so that the vapor pressure at the room temperature is reduced by 25%? Also calculate the molality of the solution obtained.

Ans: Given that,

\[{{W}_{1}}=50g\] 

\[{{M}_{1}}=18gmo{{l}^{-1}}\] 

\[{{M}_{2}}=60gmo{{l}^{-1}}\] 

The relative lowering in vapor pressure is given as,

\[\frac{{{P}^{0}}-{{P}_{S}}}{{{P}^{0}}}=\frac{{{W}_{2}}{{M}_{1}}}{{{M}_{2}}{{W}_{1}}}\] 

This is reduced to 25% which can be given as;

 \[0.25=\frac{{{W}_{2}}\times 18}{50\times 60}\] 

 \[\therefore {{W}_{2}}=41.67g\] 

Molality is given as;

\[m=\frac{{{W}_{2}}\times 1000}{{{M}_{2}}\times {{W}_{1}}}\] 

Thus,

\[m=\frac{41.67\times 1000}{60\times 50}=14m\] 

3. (a) Why is the freezing point depression considered as a colligative property?

Ans: The colligative properties depend upon the number of solute molecules in the solvent. The depression in freezing point is observed when the solute particles are introduced in the solution. Thus, it is one of the colligative properties.

(b) The cryoscopic constant of water is 1.86 k/m. Comment on this statement.

Ans: If it is said that the cryoscopic constant of water is $1.86K{{m}^{-1}}$, it means that the water has 1.86 K depression in freezing point when the solution has a molality of 1 m.

(c) Calculate the amount of ice that will separate out on cooling solution containing 50 g of ethylene glycol in 200 g ${{H}_{2}}O$  to $-9.3{}^\circ C$ . (${{K}_{f}}$ for water = $1.86Kkgmo{{l}^{-1}}$ )

Ans: Given that,

\[{{W}_{2}}=50g\] 

\[{{M}_{2}}=62gmo{{l}^{-1}}\] 

\[{{K}_{f}}=1.86Kkgmo{{l}^{-1}}\] 

\[\Delta {{T}_{f}}=9.3{}^\circ C\] 

Thus, using the formula for depression in freezing point,

\[\Delta {{T}_{f}}=\frac{{{K}_{f}}\times {{W}_{2}}\times 1000}{{{W}_{1}}\times {{M}_{2}}}\] 

\[9.3=\frac{1.86\times 50\times 1000}{{{W}_{1}}\times 62}\] 

\[\therefore {{W}_{1}}=161.29g\] 

Therefore, ice separated = 200 – 161.29 = 38.71g.

4. (a)Define osmotic pressure.

Ans: The pressure that will prevent the process of osmosis is known as osmotic pressure. It is defined as the pressure applied at the side of pure solvent to prevent it from passing through the semi – permeable membrane to the solution side by osmosis. 

(b) Why osmotic pressure is preferred over other colligative properties for the determination of molecular masses of macromolecules?

Ans: The method of osmotic pressure takes into account the molarity of the solution unlike other methods who take molality as a multiplying factor. Also, the pressure measurement takes place at room temperature during osmosis.

(c) What is the molar concentration of particles in human blood if the  osmotic pressure is 7.2 atm at normal body temperature of $37{}^\circ C$ ?

Ans: Given that,

\[\pi =7.2atm\] 

\[R=0.0821Latm{{K}^{-1}}mo{{l}^{-1}}\] 

\[T=37{}^\circ C=310K\] 

The osmotic pressure is given as;

\[\pi =CRT\] 

The molar concentration is given as;

\[C=\frac{\pi }{RT}\] 

\[C=\frac{7.2}{0.082\times 310}=0.283M\] 

Numerical Problems

1. Calculate the mass percentage of benzene (${{C}_{6}}{{H}_{6}}$ ) and carbon tetrachloride ($CC{{l}_{4}}$ ), If 22g of benzene is dissolved in 122g of carbon tetrachloride.

Ans: The mass percentage is given as;

\[\text{Mass percentage = }\frac{\text{mass of component}}{\text{mass of mixture}}\times 100\] 

Here,

Mass of benzene = 22 g

Mass of carbon tetrachloride = 122 g

Mass of mixture = 22 + 122 = 144 g

Thus,

\[\text{Mass percentage of benzene = }\frac{\text{mass of benzene}}{\text{mass of mixture}}\times 100\] 

\[=\frac{22}{144}\times 100=15.28%\] 

And,

\[\text{Mass percentage of CC}{{\text{l}}_{4}}\text{ = }\frac{\text{mass of CC}{{\text{l}}_{4}}}{\text{mass of mixture}}\times 100\] 

\[=\frac{122}{144}\times 100=84.72%\] 

2. Calculate the molarity of a solution prepared by mixing 500 mL of 2.5 M urea solution and 500 mL of 2M urea solution.

Ans: Given that,

\[{{M}_{1}}=2.5M\] 

\[{{M}_{2}}=2M\] 

\[{{V}_{1}}={{V}_{2}}=500ml\] 

Now, by the law of dilution, we get;

\[M=\frac{{{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}}\] 

\[M=\frac{\left( 2.5+2 \right)\times 500}{1000}\] 

Thus, molarity of the solution = 2.25 M.

3. The mole fraction of $C{{H}_{3}}OH$  in an aqueous solution is 0.02 and density of solution 0.994 g/cm. Determine the molality and molarity.

Ans: Given that,

Molar mass of water = 18 g/mol

Molar mass of $C{{H}_{3}}OH$= 32 g/mol

\[Density=0.994gc{{m}^{-3}}\] 

\[\text{Mole fraction of }C{{H}_{3}}OH=\frac{x}{x+y}=0.02\] 

where, x is the moles of $C{{H}_{3}}OH$ and y is the moles of water in the solution.

By calculating, we get;

\[y=49x\] 

Now, molality is given as;

\[Molality=\frac{x\times 1000}{y\times 18}=1.13m\] 

Volume of the solution is given as,

\[Volume=\frac{mass}{density}=\frac{32x+18y}{994}L\] 

Hence, molarity is given as;

\[Molarity=\frac{x\times 994}{32x+18y}=1.0875M\] 

4. 200 mL of calcium chloride solution contains $3.011\times {{10}^{22}}C{{l}^{-}}$  ions. Calculate the molarity of the solution. Assume that calcium chloride is completely ionized.

Ans: This illustration can be solved by simple mole concept;

We know that, $CaC{{l}_{2}}$ has 2 chloride ions. Thus,

The molecules of $CaC{{l}_{2}}$ will be 

\[=\frac{3.01\times {{10}^{22}}}{2}\times \frac{1}{5L}\] 

\[=7.525\times {{10}^{22}}molecules\] 

Now, we have known that a mole of a substance contains $6.022\times {{10}^{23}}molecules$ . Thus,

Molarity of solution is calculated as;

\[Molarity=\frac{7.525\times {{10}^{22}}}{6.022\times {{10}^{23}}}=0.125M\] 

5. $6\times {{10}^{-3}}$ g oxygen is dissolved per kg of sea water. Calculate the ppm of oxygen in sea water.

Ans: Given that,

\[\text{Mass of oxygen = 6}\times \text{1}{{\text{0}}^{-3}}g\] 

Mass of solution = 1 kg = 1000 g

Now,

ppm of oxygen in sea water is calculated as;

\[=\frac{\text{mass of solute}}{\text{mass of solution}}\times {{10}^{6}}\] 

\[=\frac{6\times {{10}^{-3}}\times {{10}^{6}}}{1000}=6ppm\]  

6. The solubility of oxygen in water is $1.35\times {{10}^{-3}}mol{{L}^{-1}}$at $20{}^\circ C$  and 1 atm pressure. Calculate the concentration of oxygen at $20{}^\circ C$ and 0.2 atm pressure.

Ans: Given that,

\[C=1.35\times {{10}^{-3}}mol{{L}^{-1}}\] 

P = 1 atm

We know the Henry’s law;

\[C={{K}_{H}}P\] 

Thus,

\[{{K}_{H}}=\frac{C}{P}=1.35\times {{10}^{-3}}Mat{{m}^{-1}}\] 

Now, at 0.2 atm;

\[C=1.35\times {{10}^{-3}}\times 0.2\] 

\[C=2.7\times {{10}^{-3}}M\] 

7. Two liquids X and Y on mixing form an ideal solution. The vapor pressure of the solution containing 2 mol of X and 1 mol of Y is 550 mm Hg. But when 4 mol of X and 1 mole of Y are mixed, the vapor pressure of solution thus formed is 560 mm Hg. What will be the vapor pressure of pure X and pure Y at this temperature?

Ans: Let us consider the vapor pressure of X will be ${{P}_{1}}$ and that of Y will be ${{P}_{2}}$ . Also, ${{x}_{1}}$ and ${{x}_{2}}$ be their respective mole fractions. 

Then by Raoult’s law we can say that;

Total pressure, P 

\[={{P}_{1}}{{x}_{1}}+{{P}_{2}}{{x}_{2}}\] 

 Thus, For first solution;

\[{{P}_{1}}\times \left( \frac{2}{2+1} \right)+{{P}_{2}}\times \left( \frac{1}{2+1} \right)=550\]……. 

For second solution;

\[{{P}_{1}}\times \left( \frac{4}{4+1} \right)+{{P}_{2}}\times \left( \frac{1}{4+1} \right)=560\]……. 

Now, both the equations we get,

Vapor pressure of pure component X,

\[{{P}_{1}}=575mmHg\] 

Vapor pressure of pure component Y,

\[{{P}_{2}}=500mmHg\] 

8. An aqueous solution containing 3.12 g of barium chloride in 250 g of water is found to be boil at $100.0832{}^\circ C$ . Calculate the degree of dissociation of barium chloride.

(Given molar mass $BaC{{l}_{2}}$  = $208gmo{{l}^{-1}}$, ${{K}_{b}}$  for water = 0.52 K/m)

Ans: Given that,

\[{{M}_{2}}=208gmo{{l}^{-1}}\]

\[{{K}_{b}}=0.52Kkgmo{{l}^{-1}}\] 

\[\Delta {{T}_{b}}=100.0832-100=0.0832K\] 

\[{{W}_{2}}=3.12g\] 

\[{{W}_{1}}=250g\] 

By formula we get;

\[\Delta {{T}_{b}}=i{{K}_{b}}m\] 

The Van’t Hoff factor is given as;

\[i=\frac{\Delta {{T}_{b}}}{{{K}_{b}}m}=\frac{\Delta {{T}_{b}}\times {{M}_{2}}\times {{W}_{1}}}{{{K}_{b}}\times {{W}_{2}}\times 1000}=2.67\] 

Also, we know that i = moles at equilibrium. Thus,

\[\underset{1-x}{\mathop{BaC{{l}_{2}}}}\,\rightleftarrows \underset{x}{\mathop{B{{a}^{2+}}}}\,+\underset{2x}{\mathop{2C{{l}^{-}}}}\,\] 

Solving this we get;

\[i=1-x+x+2x=2.67\] 

Therefore, the degree of dissociation is 83.5%.

9. The degree of dissociation of $Ca{{\left( N{{O}_{3}} \right)}_{2}}$  in a dilute aqueous solution, containing 7.0 g of salt per 100 g of water at $100{}^\circ C$  is 70%. If the vapor pressure of water at $100{}^\circ C$ is 760 mm of Hg, calculate the vapor pressure of the solution.

Ans: Given that,

Degree of dissociation, $\alpha $  = 70%

\[{{W}_{2}}=7g\] 

\[{{M}_{2}}=164gmo{{l}^{-1}}\] 

\[{{W}_{1}}=100g\] 

\[{{M}_{1}}=18gmo{{l}^{-1}}\] 

\[{{P}^{0}}=760mmHg\] 

Moles of $Ca{{\left( N{{O}_{3}} \right)}_{2}}$ 

\[=\frac{7}{164}=0.0426moles\] 

Moles of water 

\[=\frac{100}{18}=5.56moles\] 

Mole fraction of $Ca{{\left( N{{O}_{3}} \right)}_{2}}$, X

\[=\frac{0.0426}{0.0426+5.56}=0.00762\] 

Number of ions of $Ca{{\left( N{{O}_{3}} \right)}_{2}}$on dissociation, $n'$ = 3

The Van’t Hoff factor is given as;

\[i=\left[ 1+\left( n'-1 \right)\alpha  \right]\] 

Thus, i = 2.4

Now,

The relative lowering of vapor pressure is given as;

\[\frac{{{P}^{0}}-P}{{{P}^{0}}}=iX\] 

\[\frac{760-P}{760}=2.4\times 0.00762\] 

Thus, the vapor pressure of the solution, P = 746.1 mm Hg.

10. 2g of ${{C}_{6}}{{H}_{5}}COOH$dissolved in 25g of benzene shows depression in freezing point equal to 1.62K. Molar freezing point depression constant for benzene is $4.9Kkgmo{{l}^{-1}}$. What is the percentage association of acid if it forms a dimer in solution?

Ans: Given that,

\[{{W}_{2}}=2g\] 

\[{{W}_{1}}=25g\] 

\[{{K}_{f}}=4.9Kkgmo{{l}^{-1}}\] 

\[\Delta {{T}_{f}}=1.62K\] 

Normal molar mass of ${{C}_{6}}{{H}_{5}}COOH$ = 122 g/mol

Calculated molar mass of ${{C}_{6}}{{H}_{5}}COOH$is given as;

\[{{M}_{2}}=\frac{{{K}_{f}}\times {{W}_{2}}\times 1000}{\Delta {{T}_{f}}\times {{W}_{1}}}=241.98gmo{{l}^{-1}}\] 

The Van’t Hoff factor is given as;

\[i=\frac{\text{normal molar mass}}{\text{calc}\text{. molar mass}}=\frac{122}{241.98}\] 

Also, 

\[\underset{1-x}{\mathop{2{{C}_{6}}{{H}_{5}}COOH}}\,\rightleftarrows \underset{{}^{x}/{}_{2}}{\mathop{{{\left( {{C}_{6}}{{H}_{5}}COOH \right)}_{2}}}}\,\] 

Hence,

\[i=1-x+\frac{x}{2}=1-\frac{x}{2}=\frac{122}{241.98}\] 

Therefore, the degree of association is 99.2%.

11. Calculate the amount of NaCl which must added to one kg of water so that the freezing point is depressed by 3K. Given ${{K}_{f}}$  = $1.86Kkgmo{{l}^{-1}}$ , Atomic mass: Na = 23, Cl = 35.5).

Ans: Given that,

\[{{W}_{1}}=1kg\] 

\[{{K}_{f}}=1.86Kkgmo{{l}^{-1}}\] 

\[\Delta {{T}_{f}}=3K\] 

\[{{M}_{2}}=58.5gmo{{l}^{-1}}\] 

NaCl can dissociate by giving 2 ions. Thus, i = 2.

Therefore, by formula;

\[{{W}_{2}}=\frac{\Delta {{T}_{f}}\times {{M}_{2}}\times {{W}_{1}}}{i\times {{K}_{f}}}=47.2g\] 

Moles of NaCl 

\[=\frac{47.2}{58.5}=0.81mol\] 

12. Three molecules of a solute (A) associate in benzene to form species ${{A}_{3}}$ . Calculate the freezing point of 0.25 molal solution. The degree of association of solute A is found to be 0.8. The freezing point of benzene is $5.5{}^\circ C$  and its ${{K}_{f}}$  value is 5.13 K/m.

Ans: Given that,

3 (n’) molecules of a solute (A) associate in benzene to form species ${{A}_{3}}$ .

Molality = 0.25 m

Degree of association, $\alpha $ = 0.8

\[{{T}_{benzene}}=5.5{}^\circ C\] 

\[{{K}_{f}}=5.13K{{m}^{-1}}\] 

Now,

\[\underset{1-\alpha }{\mathop{3A}}\,\rightleftarrows \underset{{}^{\alpha }/{}_{3}}{\mathop{{{A}_{3}}}}\,\] 

\[i=1-\alpha +\frac{\alpha }{3}=1-\frac{2\alpha }{3}\] 

Van’t Hoff factor, i = 0.467

The depression in freezing point is given as;

\[\Delta {{T}_{f}}=i{{K}_{f}}m\] 

\[\Delta {{T}_{f}}=0.467\times 5.13\times 0.25=0.5985{}^\circ C\] 

The freezing point of the solution is given as;

\[T={{T}_{benzene}}-\Delta {{T}_{f}}\] 

Thus, T = 5.5 – 0.5985 = $4.90{}^\circ C$ .

13. A 5% solution of sucrose (${{C}_{12}}{{H}_{22}}{{O}_{11}}$ ) is isotonic with 0.877% solution of urea ($N{{H}_{2}}CON{{H}_{2}}$) Calculate the molecular mass of urea.

Ans: Given that,

Mass of sucrose = 5 g

Mass of urea = 0.877 g

In general, when two solutions are isotonic with each other it means they have equal osmotic pressure and due its direct proportionality with the concentrations, isotonic solutions have equal concentrations. Thus,

Conc. Of sucrose = Conc. Of urea

\[\frac{\text{mass of sucrose}}{\text{molar mass of sucrose}}=\frac{\text{mass of urea}}{\text{molar mass of urea}}\] 

We know the molar mass of sucrose as 342 g/mol. Thus, molar mass of urea is calculated as;

\[=\frac{0.877\times 342}{5}=59.9gmo{{l}^{-1}}\] 

14. Osmotic pressure of a 0.0103 molar solution of an electrolyte was found to be 0.75 atm at $27{}^\circ C$ . Calculate Van’t Hoff factor.

Ans: Given that,

\[\pi =0.75atm\] 

\[C=0.0103m\] 

\[R=0.0820Latmmo{{l}^{-1}}{{K}^{-1}}\] 

\[T=27{}^\circ C=300K\] 

By the formula of osmotic pressure, we get;

\[\pi =iCRT\] 

\[\therefore i=\frac{\pi }{CRT}=\frac{0.75}{0.0103\times 0.082\times 300}=2.95\] 

15. The maximum allowable level of nitrates in drinking water is 45 mg nitrate $ions{{m}^{-3}}$ . Express this level in ppm?

Ans: Given that,

The maximum allowable level of nitrates in drinking water = $\text{45 mg nitrates ions d}{{\text{m}}^{-3}}\text{ = 45 mg}{{\text{L}}^{-1}}$ .

We know that, $1mg{{L}^{-1}}=1ppm$ . Thus,

The maximum allowable level of nitrates in drinking water = 45 ppm.

16. 75.2 g of Phenol (${{C}_{6}}{{H}_{5}}OH$) is dissolved in 1 kg solvent of ${{K}_{f}}=14K{{m}^{-1}}$, if the depression in freezing point is 7K, then find the % of phenol that dimerises.

Ans: Given that,

\[{{W}_{2}}=75.2g\] 

\[{{M}_{2}}=94gmo{{l}^{-1}}\] 

\[\Delta {{T}_{f}}=7K\] 

\[{{K}_{f}}=14K{{m}^{-1}}\] 

\[{{W}_{1}}=1kg\] 

And,

\[\underset{1-\alpha }{\mathop{2{{C}_{6}}{{H}_{5}}OH}}\,\rightleftarrows \underset{{}^{\alpha }/{}_{2}}{\mathop{{{\left( {{C}_{6}}{{H}_{5}}OH \right)}_{2}}}}\,\] 

Thus, 

\[i=1-\alpha +\frac{\alpha }{2}=1-\frac{\alpha }{2}\] 

Now, by formula;

\[\Delta {{T}_{f}}=i{{K}_{f}}m\] 

\[\alpha =2\times \left( 1-\frac{\Delta {{T}_{f}}\times {{M}_{2}}\times {{W}_{1}}}{{{K}_{f}}\times {{W}_{2}}\times 1000} \right)\] 

\[\alpha =0.75\] 

% Phenol which will dimerise is 75%.

17. An aqueous solution of glucose boils at $100.01{}^\circ C$ . The molal boiling point elevation constant for water is $0.5Kkgmo{{l}^{-1}}$. What is the number of glucose molecule in the solution containing 100 g of water?

Ans: Given that,

\[{{W}_{1}}=100g\] 

\[{{K}_{b}}=0.5Kkgmo{{l}^{-1}}\] 

\[\Delta {{T}_{b}}=100.01-100=0.01K\] 

\[{{M}_{2}}=180gmo{{l}^{-1}}\] 

By using the formula of elevation in boiling point;

\[\Delta {{T}_{b}}={{K}_{b}}\frac{{{n}_{2}}}{{{W}_{1}}}\] 

\[\therefore {{n}_{2}}=\frac{0.01\times 0.1}{0.5}=2\times {{10}^{3}}moles\] 

From mole concept;

Number of glucose molecules, ${{N}_{a}}={{n}_{2}}\times 6.022\times {{10}^{23}}=12.046\times {{10}^{20}}molecules$ .

18. A bottle of commercial ${{H}_{2}}S{{O}_{4}}$  [density = 1.787 g/mL] is labelled as 86% by mass.

1. What is the molarity of the acid?

Ans: Given that,

Mass of acid = 86 g

Molar mass of acid = 98 g/mol

Mass of solution = 100 g

Density = 1.787 g/ml

Thus, volume is given as;

\[Volume=\frac{mass}{density}=\frac{100}{1.787}=56ml\] 

Thus, molarity is given as;

\[M=\frac{86\times 1000}{98\times 56}=15.71M\] 

2. What volume of  the acid has to be used to make 1 liter  0.2M  ${{H}_{2}}S{{O}_{4}}$?

Ans: Given that,

\[{{M}_{2}}=0.2M\] 

\[{{V}_{2}}=1L\] 

From above question,

\[{{M}_{1}}=15.71M\] 

By formula we can state that;

\[{{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}\] 

\[{{V}_{1}}=\frac{{{M}_{2}}{{V}_{2}}}{{{M}_{1}}}=0.0127L\] 

3. What is the molality of the acid?

Ans: Mass of solvent = 100 – 86 = 14 g.

Molality can be given as;

\[Molality=\frac{86\times 1000}{98\times 14}=62.68m\] 

19. A solution containing 30g of non-volatile solute exactly in 90g of water has a vapor pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapor pressure becomes 2.9 kPa at 298 K. Calculate:

1. molar mass of the solute

Ans: Given that,

\[{{W}_{2}}=30g\] 

\[{{W}_{1}}=90g\] 

\[{{P}_{S}}=2.8kPa\] 

Now, according to Raoult’s law;

\[\frac{{{P}^{0}}-{{P}_{S}}}{{{P}^{0}}}=x\approx \frac{{{W}_{2}}{{M}_{1}}}{{{M}_{2}}{{W}_{1}}}\] 

\[1-\frac{6}{{{M}_{2}}}=\frac{2.8}{{{P}^{0}}}\]……. 

After adding water,

\[{{W}_{2}}=30g\]

\[{{W}_{1}}=90+18=108g\]

      \[{{P}_{S}}=2.9kPa\]

Now, according to Raoult’s law;

\[\frac{{{P}^{0}}-{{P}_{S}}}{{{P}^{0}}}=x\approx \frac{{{W}_{2}}{{M}_{1}}}{{{M}_{2}}{{W}_{1}}}\] 

\[1-\frac{5}{{{M}_{2}}}=\frac{2.9}{{{P}^{0}}}\]……..

Now taking into account both the equations;

\[{{M}_{2}}=34gmo{{l}^{-1}}\]  

2. Vapor pressure of water at 298 K.

Ans: Now, substituting the value of molar mass in one of the equation;

\[1-\frac{6}{34}=\frac{2.8}{{{P}^{0}}}\]

Thus, ${{P}^{0}}=3.4kPa$ .

20. The vapor pressure of pure liquids A and B are 450 and 750 mm Hg respectively, at 350K. Find out the composition of the liquid mixture if total vapor pressure is 600 mm Hg. Also find the composition of the vapor phase.

Ans: Given that,

\[P_{A}^{0}=450mmHg\] 

\[P_{B}^{0}=750mmHg\] 

\[{{P}_{T}}=600mmHg\] 

Now, from Raoult’s law;

\[{{P}_{A}}=P_{A}^{0}{{x}_{A}}\] 

\[{{P}_{B}}=P_{B}^{0}{{x}_{B}}=P_{B}^{0}\left( 1-{{x}_{A}} \right)\] 

So,

\[{{P}_{T}}={{P}_{A}}+{{P}_{B}}\] 

Now,

\[{{P}_{T}}=P_{A}^{0}{{x}_{A}}+P_{B}^{0}\left( 1-{{x}_{A}} \right)\] 

\[{{x}_{A}}=\frac{{{P}_{T}}-P_{B}^{0}}{P_{A}^{0}-P_{B}^{0}}\] 

From this, 

\[{{x}_{A}}=0.5\] 

\[{{x}_{B}}=0.5\] 

This gives us;

\[{{P}_{A}}=450\times 0.5=225mmHg\] 

\[{{P}_{B}}=750\times 0.5=375mmHg\] 

Now,

Composition in vapor phase is given as;

\[{{y}_{A}}=\frac{{{P}_{A}}}{{{P}_{T}}}=\frac{225}{600}=0.375\] 

\[{{y}_{A}}=1-{{y}_{A}}=1-0.375=0.625\] 

21. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Ans: Given that,

\[{{P}_{S}}=1.004bar\] 

\[{{W}_{2}}=2g\] 

\[\text{Mass of solution = 100g}\] 

\[{{W}_{1}}=100-2=98g\] 

\[{{M}_{1}}=18gmo{{l}^{-1}}\] 

We know that VP of pure water, ${{P}^{0}}=1atm=1.013bar$ 

By Raoult’s law;

\[\frac{1.013-1.004}{1.013}=\frac{2\times 18}{{{M}_{2}}\times 98}\] 

Thus, the molar mass of the solute, ${{M}_{2}}=41.35gmo{{l}^{-1}}$ .

Important Questions of Chemistry Class 12 Solutions - Quick Review

Solution class 12 important question PDF covers the concept of solution and its types. A solution is made up of a solute and a solvent. It is a homogeneous mixture of two or more than two substances and can be classified into three types:

• Solid solutions

• Gaseous solutions

• Liquid solutions

The concentration of a solution can be described by its molarity, mole fraction, molality, and percentages. The important concepts and theorems covered in the chapter include:

• Henry’s law

• Raoult’s Law

• Ideal solutions

Henry’s law

The dissociation of gas in liquids is controlled by Henry’s law. According to this law, at a given temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas. The vapour pressure of the solvent is lowered because of the presence of a non-volatile solute.

Raoult’s Law

Raoult’s law explains the lowering of the vapour pressure of the solvent. According to this law, the relative lowering of the vapour pressure of the solvent over a solution and the mole fraction of a non-volatile solute present in the solution are equal. It can be mathematically expressed as:

Ptotal = p10 x 1 + p20 x 2

Ideal Solutions

Ideal solutions are the solutions that obey Raoult’s law over the entire range of concentrations. Raoult’s law has two types of deviations:

• Positive deviations

• Negative deviations

Other important concepts covered by the important questions include mole fraction, the role of molecular interaction in a solution of water and alcohol, the characteristics of a solution, types of solutions, examples of ‘liquid in solid’ solutions, Can’t Hoff factor, reverse osmosis and its uses, cryoscopic constant (Molal freezing point depression) application of Henry’s law, application of Raoult’s Law and the concept of molarity.

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Related Study Materials for Class 12 Chemistry Chapter 1

CBSE Class 12 Chemistry Chapter-wise Important Questions

CBSE Class 12 Chemistry Chapter-wise Important Questions and Answers cover topics from other chapters, helping students prepare thoroughly by focusing on key topics for easier revision.

Additional Study Materials for Class 12 Chemistry

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