Solutions Class 12 Chemistry Exam Questions

Please refer to the below Solutions important questions for Class 12 Chemistry. These questions and answers have been prepared as per the latest NCERT Book for Class 12 Chemistry. Students should go through chapter wise Class 12 Chemistry Important Questions designed as per the latest examination pattern issued by CBSE.

Very Short Answer Questions :

Question. Define Raoult’s law.  
Answer : Raoult’s law : For a solution of volatile liquids, the partial pressure of each component in the  solution is directly proportional to its mole fraction. Thus, for any component, partial vapour pressure, p ∝ x ⇒ p = p°. x
where p° = vapour pressure of pure component 
          x = mole fraction of that component

Question. State the condition resulting in reverse osmosis.  
Answer : Reverse osmosis occurs when a pressure larger than the osmotic pressure is applied to the solution side.

Question. Define the following term :
Azeotrope  
Answer : Azeotropes are the binary mixtures of solutions that have the same composition in liquid and vapour phases and that have constant boiling points.
It is not possible to separate the components of azeotropes by fractional distillation.

Question. State the following :
Raoult’s law in its general form in reference to solutions.  
Answer : Raoult’s law : For a solution of volatile liquids, the partial pressure of each component in the  solution is directly proportional to its mole fraction. Thus, for any component, partial vapour pressure, p ∝ x ⇒ p = p°. x
where p° = vapour pressure of pure component 
          x = mole fraction of that component

Question. Define the following term :
Molality (m)
Answer : Molality (m) is defined as the number of moles of the solute per kilogram of the slovent and is  expressed as :
Molality (m) = Moles of solute/Mass of solvent (in kg)

Question. Define the following term :
Colligative properties
Answer : Properties which depend upon the number of solute particles irrespective of their nature relative to the total number of particles present in the solution are called colligative properties.

Question. Define the following term :
Molarity (M)
Answer : Molarity : It is the number of moles of the solute dissolved per litre of the solution. It is denoted by M.
∴ Molarity = Moles of solute/Volume of solution in litre

Question. Explain the following :
Henry’s law about dissolution of a gas in a liquid.  
Answer : Henry’s law states that, the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.
p = K_H⋅x where, K_H = Henry’s law constant. Different gases have different K_H values at the same temperature.

Question. In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes?  
Answer : Non-ideal solutions that show negative deviation from Raoult’s law form maximum boiling azeotropes.

Short Answer Questions :

Question. A 0.1539 molal aqueous solution of cane sugar (mol. mass = 342 g mol^–1) has a freezing point of 271 K while the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution containing 5 g of glucose (mol. mass = 180 g mol^–1) per 100 g of solution.  
Answer : 

Question. Explain why aquatic species are more comfortable in cold water rather than in warm water.  
Answer : Increase in temperature decreases the solubility of oxygen in water. As a result, amount of dissolved oxygen decreases. It becomes more diffcult to breathe as oxygen is less. Hence, the aquatic species are not comfortable in warm water.

Question. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase. 
Answer :

Question. Define :
(i) Mole fraction
(ii) Molality 
Answer : (i) Mole fraction is the ratio of number of moles of solute or solvent and total number of moles of solution. It is denoted by x.
x_solute = n₂/n₁+n₂, x_solvent = n₁/n₁+n₂
(ii) Molality of a solution can be defined as the number of moles of solute dissolved in one kg solvent. It is denoted by m. 
m = Number of moles of solute/Mass of solvent in kg = n₂/W₁

Question. A solution of glucose (C₆H₁₂O₆) in water is labelled as 10% by weight. What would be the molality of the solution?
(Molar mass of glucose = 180 g mol^–1) 
Answer : 

Question. Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing
(i) 1.2% sodium chloride solution?
(ii) 0.4% sodium chloride solution? 
Answer : (i) 1.2% sodium chloride solution is hypertonic with respect to 0.9% sodium chloride solution or  blood cells thus, on placing blood cells in this solution exosmosis takes place that results in shrinking of  cells.
(b) 0.4% sodium chloride solution is hypotonic with respect to 0.9% sodium chloride solution or blood cells thus, on placing blood cells in this solution endosmosis takes place that results in swelling of cells.

Question. State Raoult’s law. How is it formulated for solutions of non-volatile solutes?  
Answer : Raoult’s law : For a solution of volatile liquids, the partial pressure of each component in the solution is directly proportional to its mole fraction. Thus, for any component, partial vapour pressure, p ∝ x ⇒ p = p°. x
where p° = vapour pressure of pure component 
          x = mole fraction of that component
Raoult’s law for solution containing non-volatile solute : It states that partial vapour pressure of a solution of non-volatile solute, p_solution is directly proportional to the mole fraction of the solvent in the solution.
Mathematically,
p_solution = p°_solvent × x_solvent
where p°solvent = vapour pressure of the pure solvent at the given temperature.
or, p° − psol/p° =x₂

Question. A solution containing 8 g of a substance in 100 g of diethyl ether boils at 36.86°C, whereas pure ether boils at 35.60°C. Determine the molecular mass of the solute. (For ether K_b = 2.02 K kg mol^–1) 
Answer : 

Question. Why does a solution containing non-volatile solute have higher boiling point than the pure solvent? Why is elevation of boiling point a colligative property? 
Answer : The boiling point of the solution is always higher than that of the pure solvent. As the vapour pressure of the solution is lower than that of the pure solvent and vapour pressure increases with increase in temperature. Hence, the solution has to be heated more to make the vapour pressure equal to the atmospheric pressure.
Elevation of boiling point is a colligative property because it depends on number of solute particles present in a solution.

Question. Define the following term :
Ideal solution
Answer : A solution which obeys Raoult’s law of vapour pressure for all compositions is called ideal solution.
In this solution ΔV_mix = 0, ΔH_mix = 0
A ……. B interaction = A ….. A and B ….. B interactions.

Question. 30 g of urea (M = 60 g mol^–1] is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.
Answer :

Question. Write two differences between ideal solutions and non-ideal solutions.
Answer : The two differences between ideal solutions and non ideal solutions are as follows :
(i) In ideal solutions ΔV_mixing = 0 and ΔH_mix= 0 whereas in non ideal solutions, ΔV_mix ≠ 0 and ΔV_mix ≠ 0.
(ii) In ideal solutions, each component obeys Raoult’s law at all temperatures and concentrations whereas in non ideal solutions, they do not obey Raoult’s law.

Question. A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K.
(Given : Molar mass of sucrose = 342 g mol^–1, molar mass of glucose = 180 g mol^–1)
Answer : Molality (m) of sucrose solution

Question. Define the following terms :
(i) Abnormal molar mass
(ii) van’t Hoff factor (i)
Answer : (i) The molar mass which is either lower or higher than the expected or normal value is known as abnormal molar mass.
(ii) van’t Hoff factor (i) is defined as the ratio of the experimental (observed) value of colligative property to the calculated value of colligative property.
     i = Observed colligative property/Calculated colligative property
or, i = Normal molar mass/Abnormal molar mass

Question. Calculate the mass of compound (molar mass = 256 g mol^–1) to be dissolved in 75 g of  benzene to lower its freezing point by 0.48 K.
(K_f = 5.12 K kg mol^–1).  
Answer : 

Question. Differentiate between molarity and molality in a solution. What is the effect of temperature  change on molarity and molality in a solution? 
Answer : 

Question. State Henry’s law and mention two of its important applications.  
Answer : Henry’s law states that, the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.
p = K_H⋅x where, K_H = Henry’s law constant. Different gases have different K_H values at the same  temperature.
Applications of Henry’s law :
(i) To increase the solubility of CO₂ in soda drinks and soda water, the bottle is sealed under high pressure.
(ii) To minimise the painful effects of decompression sickness in deep sea divers, oxygen diluted with less  soluble helium gas is used as breathing gas.

Question. (i) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?
(ii) What happens when the external pressure applied becomes more than the osmotic pressure of solution? 
Answer : (i) The elevation in boiling point of a solution is a colligative property which depends on the number of moles of solute added. Higher the concentration of solute added, higher will be the elevation in boiling point. Thus, 2 M glucose has higher boiling point than 1 M glucose solution.
(ii) When the external pressure applied becomes more than the osmotic pressure of solution then the solvent molecules from the solution pass through the semipermeable membrane to the solvent side and the process is called reverse osmosis.

Question. Calculate the mass of NaCl (molar = 58.5 g mol^–1) to be dissolved in 37.2 g of water to lower the freezing point by 2°C, assuming that NaCl undergoes complete dissociation.
(K_f for water = 1.86 K kg mol^–1)  
Answer : 

Question. An aqueous solution containing 12.48 g of barium chloride in 1.0 kg of water boils at 373.0832 K. Calculate the degree of dissociation of barium chloride.
[Given K_b for H₂O = 0.52 K m^–1; Molar mass of BaCl₂ = 208.34 g mol^–1] 
Answer : 

Question. Find the boiling point of a solution containing 0.520 g of glucose (C₆H₁₂O₆) dissolved in 80.2 g of water.
[Given : K_b for water = 0.52 K/m] 
Answer :

Question. What mass of NaCl must be dissolved in 65.0 g of water to lower the freezing point of water by 7.50°C? The freezing point depression constant (K_f) for water is 1.86°C/m. Assume van’t Hoff factor for NaCl is 1.87. 
(Molar mass of NaCl = 58.5 g mol^–1).  
Answer : 

Question. Outer hard shells of two eggs are removed. One of the egg is placed in pure water and the other is placed in saturated solution of sodium chloride. What will be observed and why? 
Answer : The egg placed in pure water will swell because the concentration of proteins is high inside the egg as compared to water. Therefore, endosmosis occurs and water diffuses through the semipermeable membrane. The egg which is placed in sodium chloride solution will shrink due to osmosis of water out of the egg.

Question. Calculate the boiling point of one molar aqueous solution (density 1.06 g mL^–1) of KBr.
[Given : K_b for H₂O = 0.52 K kg mol^–1, atomic mass : K = 39, Br = 80]  
Answer : 

Question. A decimolar solution of potassium ferrocyanide K₄[Fe(CN)₆] is 50% dissociated at 300 K.
Calculate the value of van’t Hoff factor for potassium ferrocyanide.  
Answer : 

Question. Define the terms, ‘osmosis’ and ‘osmotic pressure’. What is the advantage of using osmotic pressure as compared to other colligative properties for the determination of molar masses of solutes in solutions?  
Answer : Osmosis : The spontaneous movement of the solvent molecules from the pure solvent or from a dilute solution to a concentrated solution through a semi-permeable membrane is called osmosis.
Osmotic Pressure : The minimum excess pressure that has to be applied on the solution to prevent the passage of solvent molecules into it through semipermeable membrane is called osmotic pressure.
Osmotic pressure is a colligative property because it depends on the number of solute particles and not on their nature.
The osmotic pressure method has the advantage over other methods because
(i) osmotic pressure can be measured at room temperature and the molarity of the solution is used instead of molality. 
(ii) its magnitude is large as compared to other colligative properties even for very dilute solutions.

Question. The molecular masses of polymers are determined by osmotic pressure method and not by measuring other colligative properties. Give two reasons.  
Answer : The egg placed in pure water will swell because the concentration of proteins is high inside the egg as compared to water. Therefore, endosmosis occurs and water diffuses through the semipermeable membrane. The egg which is placed in sodium chloride solution will shrink due to osmosis of water out of the egg.

Question. Calculate the freezing point of solution when 1.9 g of MgCl₂(M = 95 g mol^–1) was dissolved in 50 g of water, assuming MgCl₂ undergoes complete ionization.
(K_f for water = 1.86 K kg mol^–1)  
Answer :

Question. Define azeotropes. What type of azeotrope is formed by negative deviation from Raoult’s  law? Give an example.  
Answer : Azeotropes are the binary mixtures of solutions that have the same composition in liquid and vapour phases and that have constant boiling points.
It is not possible to separate the components of azeotropes by fractional distillation.
A maximum boiling azeotrope is formed by solutions showing a large negative deviation from Raoult’s law at a specific compostion.
For example Chloroform – acetone mixture.

Question. The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g of benzene is lowered by 0.45°C. Calculate.
(i) the molar mass of acetic acid from this data
(ii) van’t Hoff factor
[For benzene, K_f = 5.12 K kg mol^–1]
What conclusion can you draw from the value of van’t Hoff factor obtained? 
Answer :