2nd PUC Chemistry Question Bank Chapter 2 Solutions

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Karnataka 2nd PUC Chemistry Question Bank Chapter 2 Solutions

Question 1.
What is binary solution?
Answer:
A solution which consists of two components is called binary solution.

Question 2.
What is saturated solution?
Answer:
A solution in which no more solute can be dissolved at the same temperature and pressure is called a saturated solution.

Question 3.
Define the terra molarity.
Answer:
It is defined as the number of moles of solute present in one liter of its solution.

Question 4.
How does molarity changes with temperature?
Answer:
Molality decreases with increase in temperature.

Question 5.
Define molality.
Answer:
Molality is defined as the number of moles of the solute present in one kilogram of solvent.

Welcome to our step-by-step math solver Partial fraction calculator.

Question 6.
Define mole fraction.
Answer:

Question 7.
State Henry’s law. Give its mathematical form.
Answer:
At constant temperature the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of the solution.
P = K_H . x
p : partial pressure of the gas.
x : mole fraction of the gas.
K_H : Henry’s constant.

Question 8.
Give the relation between solubility of gas in liquid and Henry’s constants
Answer:
Higher the value of K_H at a given pressure the lower is the solubility of the gas in the liquid.

Question 9.
At given temperature and pressure nitrogen gas is more soluble in water than helium gas. Which one of them has higher K_H value.
Answer:
Helium gas.

Question 10.
k_H (Henry’s constant) value for H₂ and O₂ 69.16 k bar; 34.86 k bar respectively at 293K. Which one of them is highly soluble in water.
Answer:
Oxygen gas.

Question 11.
Give three applications of Henry’s law.
Answer:
1. To increase the solubility of carbon dioxide, in soft drinks the bottles are sealed under high pressure.

2. To avoid the toxic effects of high concentration of nitrogen in the blood i.e., to avoid the disease bends, sea divers carry oxygen cylinder diluted with helium.

3. Due to lower partial pressure of oxygen at high altitude, concentration of oxygen decreases in the blood tissues of mountain climber’s. This causes climber’s to become weak and unable to think clearly.

Question 12.
Name the law behind the dissolution of CO₂ gas in soft drinks under high pressure.
Answer:
Henry’s Law.

Question 13.
What happen’s to the solubility of gas in liquid with the increase in temperature? Give reason.
Answer:
Dissolution of gas in a liquid is an exothermic process and hence on increasing temperature, solubility decreases.

Question 14.
What is the effect of increase in pressure on the solubility of a gas in a liquid.
Answer:
Increase in pressure increases the solubility of a gas in a liquid.

Question 15.
How solubility of gas in liquid varies with
(i) Temperature
(ii) Pressure.
Answer:
Solubility of gas decreases with increase in temperature and increases with increase in pressure.

Question 16.
Aquatic animals are more comfortable in cold water rather than in warm water. Give reason.
Answer:
Because solubility of oxygen is more in cold water than in warm water.

Question 17.
Two liquids A and B boils at 145°C and 190°T respectively. which of them has higher vapour pressure at 80°C.
Answer:
Liquid A has higher vapour pressure.

Question 18.
State Raoult’s law of liquid mixtures.
Answer:
The partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
P₁ α X₁ p₁ = P₁⁰ x₁
P₁ → Partial vapour pressure of the component
P₁⁰ → Vapour pressure of pure component
x₁ → Mole fraction of the component in the solution.

Question 19.
What are ideal solutions?
Answer:
The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions.

Question 20.
Give two characteristics of an ideal solution.
Answer:

There is no change in enthalpy when an ideal solution is formed from their components, i.e. ΔH_mixing = 0.
There is no change in volume when an ideal solution is formed from their components, i.e. ΔV_mixing = 0.

Question 21.
Give three examples for an ideal solutions.
Answer:

Mixture of n-hexane and n-heptane.
Mixture of bromoethane and chloroethane.
Mixture of benzene and toluene.

Question 22.
What are non-ideal solutions? (Jive two characteristics.
Answer:
The solutions which do not obey Raoult’s law, over the entire range of concentration are called non-ideal solutions.

There is a change in enthalpy when a non-ideal solution formed from their components, i.e. ΔH_mix ≠ 0.
There is a change in volume when a non-ideal solution formed from their components, i.e. ΔV_mix ≠ 0.

Question 23.
What is meant by non-ideal solution showing the positive deviation from Raoults law?
Answer:
If the vapour pressure of solution is greater than the expected from Raoult’s law, then that solution is called non-ideal solution showing positive deviation from Raoult’s law.

Question 24.
What is meant by non-ideal solution showing -ve deviation from Raoults law?
Answer:
If the vapour pressure of a solution is less than the expected from Raoult’s law, then that solution is called non-ideal solution showing negative deviation from Raoult’s law.

Question 25.
Givc examples for non-ideal solution showing +ve deviation from Raoult’s law.
Answer:

Mixture of ethanol and acetone.
Mixture of carbon disulphide and acetone.

Question 26.
Give examples for non-ideal solution showing negative deviation from Raoult’s law.
Answer:

Mixture of phenol and aniline.
Mixture of chloroform and acetone.

Question 27.
Mention the enthalpy of mixing (ΔH_mix) value to form an ideal solution.
Answer:
ΔH_mix = 0

Question 28.
On mixing equal volumes of ethanol and acetone what type of deviation from Raoult’s law is expected?
Answer:
+ve deviation.

Question 29.
Mention the differences between ideal and non-ideal solutions.
Answer:

Ideal Solution	Non Ideal Solution
1. Obey’s Raoult’s law over the entire range of concentration.	1. Doesn’t Obey’s Raoult’s law over the entire range of concentration.
2. There is no change in enthalpy when the solution is formed from their components. ΔH_mix = 0	2. There is no change in enthalpy when the solution is formed from their components. ΔH_mix ≠ 0
3. There is no change in volume when the solution is formed from their components. ΔV_mix = 0	3. There is no change in volume when the solution is formed from their components. ΔV_mix ≠ 0
4. The interactions between the components in solutions are similar to those in the pure components.	4. The interactions between the components in solutions are similar to those in the pure components.

Question 30.
What is azeotropic mixture? Give example.
Answer:
Binary mixture boils at the same temperature without undergoing any change in composition is known as azeotropic mixture.
Example: Mixture of ethanol and water.

Question 31.
What is minimum boiling azeotrope? Give example.
Answer:
The solution at a specific composition have constant boiling point less than the either of the two pure components is called minimum boiling azeotrope.
Example: 95% ethanol and 5% water forms minimum boiling azeotrope at constant temperature of 78°C.

Question 32.
What is maximum boiling azeotrope? Give example.
Answer:
The solution at a specific composition have constant boiling point higher than the either of the two pure components is called maximum boiling azeotrope.
Example: 68% nitric acid and 32% water forms maximum boiling azeotrope at constant temperature of 120.5 C.

Question 33.
Components of a non-ideal binary solution cannot be completely separated by fractional distillation why?
Answer:
It forms a azetrope.

Question 34.
What type of non-ideal solution shows minimum boiling azeotrope?
Answer:
Non-ideal solution with positive deviation forms minimum boiling azeotrope.

Question 35.
What type of non-ideal solution shows maximum boiling azeotrope?
Answer:
Non-ideal solution with negative deviation forms maximum boiling azeotrope.

Question 36.
10ml of liquid A is mixed with 10ml of liquid B, the volume of the resultant solution is 19.9ml. What type of deviation expected from Raoult’s law.
Answer:
Negative or – ve.

Question 37.
What are colligative properties?
Answer:
Properties of the dilute solutions which depend only on the number of solute particles but not on the nature of the solute particles-are called colligative properties.

Question 38.
Name four colligative properties.
Answer:
The four important colligative properties are:

Relative lowering of vapour pressure.
Elevation in boiling point.
Depression in freezing point.
Osmotic pressure.

Question 39.
On what factor the value of colligative property depends?
Answer:
Colligative property depends upon the number of solute particles.

Question 40.
State Raoult’s law of a solutions containing nonvo1atì1e solute Give its mathematical expression for the law.
Answer:
Relative lowering of vapour pressure of a solution containing non volatile solute is equal to mole fraction of the solute.

p° → Vapour pressure of pure solvent
n₁ → No. of moles of solvent
p → Vapour pressure of solution
n₂ → No. of moles of solute.

Question 41.
Give equation to calculate molar mass of a solute using relative lowering of vapour pressure.
Answer:

M₂ → Molar mass of solute
W₂ → Mass of solute in gramš
W₁ → Mass of solvent in grams
M₁ → Molar mass of solvent
p° → Vapour pressure of solvent
p → Vapour pressure of solution.

Question 42.
What happens to vapour pressure of water if a table spoon of sugar is added to it?
Answer:
Vapour pressure decreases

Question 43.
Among sea water and river water which has high vapour pressure.
Answer:
River water.

Question 44.
Name two factors on which vapour pressure of a solvent depends.
Answer:

Temperature
Nature of solvent.

Question 45.
Explain the elevation in boiling point of solution.
Answer:
When a non-volatile solute is added to solvent vapour pressure of a solution decreases.
This solution has higher boiling point than the boiling point of solvent.
For a dilute solution elevation in boiling point (ΔT_b) is directly proportional to the molal concentration of the solute in a solution.

ΔT_b : is elevation in boiling point
i.e., ΔT_b = T_b – T⁰_b
T_b : is boiling point of solution
T⁰_b : is boiling point of pure solvent
K_b : is motal elevation constant
m : is malality of a solution.

Question 46.
Give equation to calculate molar mass of a solute based on elevation in boiling point temperature.
Answer:

M₂ = Molar mass of solute
K_b = Molal elevation constant. It’s value for water at 373K is 0.52 K kg mol^-1.
W_b = Mass of solute in grams
W₁ = Mass of solvent in grams
ΔT_b = Elevation in boiling point.

Question 47.
Explain the depression in freezing point of a solution.
Answer:
When a non-volatile solute is added to solvent vapour pressure of a solution decreases.
This solution has lower freezing point than the freezing point of pure solvent.
For a dilute solution depression of freezing point (ΔT_t) is directly proportion to molality (m) of the solution.
ΔT_f α m
ΔT_f = K_fm
ΔT_f is depression in freezing point, i.e., ΔT_f = ΔT⁰_f – T_f
T_f is freezing point of solution
T_f⁰ is freezing point of solvent.

Question 48.
Give equation to calculate molar mass of a solute based on depression in freezing point temperature.
Answer:

M₂ = molar masss of solute.
K_f = molal depression constant. It’s value for water at 273 K is 1.86 K.kg mol^-1.

Question 49.
Give the unit for molal boiling point elevation constant and molal freezing point depression constant.
Answer:
For both molal boiling point elevation constant (K_b) and molal freezing point depression constant (K_f) unit is same, i.e. K kg mol^-1.

Question 50.
What is osmosis?
Answer:
The phenomenon of the flow of solvent molecules through a semipermeable membrane from solvent, side to solution side is called osmosis.

Question 51.
What is osmotic pressure?
Answer:
The external pressure applied on solution side to prevent the osmosis of the solvent molecules separated by semipermeable membrane is called osmotic pressure.

Question 52.
Give the relation between osmotic pressure and concentration of a solution.
Answer:
Osmotic pressure of a solution is directly proportional to molarity (c) of the solution at a given temperature. π = CRT
n → Osmotic pressure
C → Molarity of a solution
R → Gas constant it’s value is 0.082 L atm mol^-1
T → Absolute temperature.

Question 53.
Give equation to calculate molecular mass of a solute using osmotic pressure method.
Answer:
M₂ → Molecular mass of a solute.
R → gas constant.
T → Absolute temperature.
W₂ → Mass of solute in grams.

Question 54.
Give three advantages of determination of molecular mass by osomtic pressure method compared to other colligative methods.
Answer:

Osmotic pressure measurement is around the room temperature.
Molarity of a solution is used instead of molality.
Osmotic pressure (π) magnitude is large even for very dilute solutions.

Question 55.
What advantage the osmotic pressure method has over the other colligative methods for the determination of molecular mass of macro molecules?
Answer:
Stability of macromolecules depends upon the temperature. The osmotic pressure method has the advantage over other methods as pressure measurement is around the room temperature.

Question 56.
What are isotonic solutions? Give example.
Answer:
Two solutions having same osmotic pressure at a given temperature are called isotonic solutions. Example: Blood cells are isotonic with 0.9 (m/v) sodium chloride solution (saline).

Question 57.
What is hypertonic solution? Give example.
Answer:
The solution whose osmotic pressure is less than that of the other is called hypertonic solution.
Example: Blood cells suspended in a solution containing more than 0.9% (m/v) sodium chloride solution (saline).
In this solution, water will flow out the blood cells and they would shrink.

Question 58.
What is hypotonic solution? Give example.
Answer:
The solution whose osmotic pressure is more than that of the other is called hypotonic solution. Example: Blood cells suspended in a solution containing less than 0.9% (m/v) sodium chloride (saline) solution. In this solution, water will flow into the cells and they would swell.

Question 59.
What is reverse osmosis? Mention one of its practical utility.
Answer:
The direction of osmosis can be reversed when a pressure larger than the osmotic pressure is applied to the solution side. This phenomenon is called reverse osmosis.
Reverse osmosis is used in desalination of sea water.

Question 60.
What will happcn’s if pressure greater than osmotic pressure is applied on the solution separated by semipermeable membrane from the solvent.
Answer:
Reverse osmosis takes place.

Question 61.
Explain the process of desalination of sea water.
Answer:

Sea water is separated from fresh water by cellulose acetate semipermeable membrane. When pressure more than osmotic pressure is applied on sea water outlet side pure water squeezed out of the sea water through the membrane due to reverse osmosis.

Question 62.
What is van’t Hoff factor?
Answer:

van’t Hoff factor (i) gives the extent of dissociation or association of solute in a given solvent.

Question 63.
Under what conditions van’t Hoff factor ‘i’ is
1. greater than 1
2. less than 1
3. equal to 1
Answer:
1. When the solute under goes dissociation.
2. When the solute under goes association.
3. The solute does not under goes association or dissociation.

Question 64.
What docs the van’t Hoff factor ‘i’ for a solute in a solvent account for?
Answer:
Van’t Hoff factor (i) gives the extent of dissociation or association of a solute in a given solvent.

Question 65.
Van’t Hoff factor ‘i! is greater than one for aqueous solution containing potassium chloride why?
Answer:
Potassium chloride undergoes dissociation.

Question 66.
What is the van’t Hoff factor of a completely ionised aqueous solution of
1. NaCI
2. KCl
3. MgSO₄
4. K₂SO₄
Answer:
1. NaCl → Na⁺ + Cl^–
i = \(\frac { 2 }{ 1 }\) = 2

2. MgSO₄ → Mg²⁺ + SO₄^2-
i = \(\frac { 2 }{ 1 }\) = 2

3. KCl → k⁺ + Cl^–
i = \(\frac { 2 }{ 1 }\) = 2

4. K + SO₄^2-
i = \(\frac { 3 }{ 1 }\) = 3

Question 67.
What is the van’t Hoff factor of acetic acid when it dimirises in benzene.
Answer:
i = \(\frac { 1 }{ 2 }\) = 0.5

2nd PUC Chemistry Solutions Problems and Solutions

Determination of molecular mass of a solute

(a) Relative lowering of vapour pressure method.

Question 1.
The vapour pressure of a pure benzene at certain temperature is 0.850 bar. A non-volatile solid weighing 0.5 g when added to 39 grams of benzene (molar mass 78) vapour pressure of the solution then is 0.845 bar. What is the molar mass of the substance?
Answer:
Data:
Vapour pressure of solvent benzene = P° = 0.850 bar.
Vapour pessure of solution = P = 0.845 bar.
Lowering of vapour pressure = P° – P = 0.850 – 0.845 = 0.005 bar.
\(\frac{\mathrm{P}^{0}-\mathrm{P}}{\mathrm{P}^{0}}=\frac{0.005}{0.850}\) = 0.00588
Mass of solute = W₂ = 0.5 g;
Mass of solvent benzene = W₁ = 39 g
Molar mass of benzene = 78 g mol^-1

Question 2.
5.8 g of a non-volatile solute was dissolved in 100g of carbon disulphide (CS₂). The vapour pressure of the solution was found to be 190 mm of Hg. Calculate the molar mass of the solute. Given the vapour pressure of pure CS₂ is 195 mm of Hg (Molar mass of CS₂ = 76 g mol^-1).
Answer:
Data: Vapour pressure of solvent = P° = 195 mm
Vapour pressure of solution = P = 190 mm
P°- P = 195 – 190 = 5mm
Relative lowering of vapour pressure = \(\frac{\mathrm{P}^{\circ}-\mathrm{P}}{\mathrm{P}}=\frac{5}{195}\) = 0.02564
Mass of solute in grams = W₂ = 5.8g
Mass of solute (CS₂) in grams = 100g
Molar mass of solvent (CS₂) = 76g mol^-1

Question 3.
The vapour pressure of pure benzene at a certain temperautre is 200 mm Hg. The vapour pressure of a solution containing 2 g of a non¬volatile solute in 78 g of benzene is 195 mm Hg. What is the molar mass of the solute? (Molar mass of benzene 78 g mol^-1).
Answer:
80 g/mol.

Question 4.
Aqueous solution of 3% non-volatile solute exerts a pressure of 1 bar at the boiling point of the solvent. Calculate the molar mass of solute.
Answer:
Data:
Vapour pressure of water at its [boiling point temperature]. = P°= 1.013 bar
Vapour pressure of solution at normal boiling point P = 1 bar 3% solution = 3 g of solute in 100 g of solution.
Mass of solute = 3 g
Mass of solvent = 100 – 3 = 97 g
Relativc lowering of vapour pressure = \(\frac{P^{\circ}-P}{P}=\frac{1.013-1}{1.013}\) = 0.0128
Molar mass of water = 18 g mol^-1

(b) Elevation in boiling point method

Question 5.
On dissolving 2.34g of solute in 10 g of benzene, the boiling point of the solution was higher than that of the benzene by 0.81 K.K_b value for benzene is 2.53 K. kg mol^-1. Calculate the molar mass of the solute.
Answer:
Data: Mass of solute = W₂ = 2.34 g;
Mass of solvent benzene = W₁ = 40 g
K_b = 2.53 k.kg mol^-1
Elevation in boiling point = 0.81 k
Molar mass of solute = M₂ = ?

Question 6.
The boiling point of benzene is 353.23 K. When 1.80 g of non-volatile, non-ionizing solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11K. Calculate the molar mass of the solute. [K_b value for benzene is 2.53 K. kgmol^-1].
Answer:
Boiling point of solution (T) = 354.11 K.
Boiling point of solvent (T°) = 353.23 K.
Elevation in boiling point (ΔT_b) = T – T° = 354.11 – 353.23 = 0.88 K.
Mass of solute = W₂ = 1.80 g;
Mass of solvent = W₁ = 90 g
K_b = 2.53 K.kg mol^-1
Molar mass of solute = M₂ = ?

Question 7.
A solution containing 0.564 g of solute in 36.8 g of acetone boils at 56.46°C. Calculate the molar mass of solute if the boiling point of acetone is 56.30°C. [K_b for acetone 1.92 K kg.mol^-1].
Answer:
183.9 g mol^-1

Question 8.
Chloroform boils at 61.7° C and the solution of 6.21 g of an organic compound in 24 g of chloroform boils at 68.04°C. What would be the molar mass of compound if K_b value is 3.63 K kg mol^-1.
Answer:
148.15 g mol^-1

Question 9.
A solution containing 18 g of non-volatile, non-electrolyte is dissolved in 200 g of water freezes at 272.07 K. Calculate the molar mass of solute. [Given K_f = 1.86 K kg mol^-1].
Answer:
Freezing point of water (T°) = 273 K;
Freezing point of solution (T) = 272.07 K
Depression in freezing point ΔT_f = (T° – T) = 273 – 272.07 = 0.93 K
Mass of solute = W₂ = 18 g;
Mass of solvent = W₁ = 200 g
Depression = K_f = 1.86 K. kg mol^-1

Question 10.
1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K. kg mol^-1. Find the molar mass of the solute.
Answer:
Depression in freezing point = ΔT_b = 0.4 K
Mass of solute = W₂ = 1g
Mass of solvent = W₁ = 50 g
Depression constant = K_f = 5.12 K. kg mol^-1

Question 11.
15.0 g of unknown substance was dissolved in 450 g of water, the resulting solution was found to freeze at -0.34°C. Calculate the molar mass of the substance [K_f for water = 1.86 K kg mol^-1].
Answer:
182.35 g mol^-1

Question 12.
5% solution of a substance in water has freezing point 269.06 K. Calculate molar mass of solute. Freezing point of pure water 273.15 K.
Answer:
180 g mol^-1 [K_f = 14 K. kg mol^-1].

(d) Osmotic pressure method:

Question 13.
200 cm³ of an aqueous solution of a protein contains 1.26 g of protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10^-3 bar. Calculate the molar mass of the protein. (R = 0.083 L bar mol^-1 K^-1).
Answer:
Data:
Volume of the solution = V = 200 cm³ = 0.2 L
Mass of solute = W₂ = 1.26 g
Temperature = T = 300 K;
Osmotic pressure = p = 2.57 × 10^-3 bar
R = 0.083 L. bar mol^-1 K^-1;
Molar mass = M₂ = ?

Question 14.
300 cm³ of an aqueous solution of a protein contains 2.12 g of the protein, the osmotic pressure of such a solution at 300 K is found to be 3.89 × 10^-3 bar. Calculate the molar mass of protein. (R = 0.0823 L bar mol^-1 K^-1).
Answer:
Data: Volume of the solution = V = 300cm³ = 0.3 L
Mass of solute = W₂ = 2.12 g;
Temperature = T = 300 K
Osmotic pressure = π = 3.80 × 10^-3 bar;
R = 0.0823 L bar K^-1 mol^-1
Molecular mass = M₂ = ?

Question 15.
3 gram of non-volatile solute in a 1000 cm3 of water shows an osmotic pressure of 2 bar at 300 K. Calculate the molar mass of the solute (R = 0.083 L bar K^-1 mol^-1).
Answer:
37.35 g mol^-1

Question 16.
Osmotic pressure of a solution containing 7 g of a protein present in deciliter of a solution is 3.3 × 10^-2 bar at 37° C. Calculate the molar mass of protein.
Answer:
54579 g mol^-1

Type – 2 Calculation of vapour pressure and osmotic pressure of a solution

Question 17.
The vapour pressure of ethyl alcohol at 298 K is 40 mm of Hg. Its mole fraction in a solution with methyl alcohol is 0.80. What is its vapour pressure in solution, if the mixture obeys Raoult’s law.
Answer:
Date:
Vapour pressure of ethyl alcohol in solution = P₁ = ?
Vapour pressure of pure ethyl alcohol = P° = 40 mm Hg
Mole fraction of ethyl alcohol = x₁ = 0.8
According to Raoult’s law = P₁ = P₁⁰ × x
P₁ = 40 × 0.8 = 32 mm Hg.

Question 18.
Vapour pressure of chloroform and dichloroethane at 298 K are 200 mm Hg and 415 mm Hg respectively. If the mole fraction of chloroform is 0.312. Calculate vapour pressure of the solution.
Answer:
Data: Vapour pressure of chloroform P₁⁰ = 200 mm Hg
Vapour pressure of dichloroethane P₂⁰ = 415 mm Hg
Mole fraction of chloroform n₁ = 0.312
Mole fraction of dichloroethane = n₂ = 1 – 0.312 = 0.6878
P_T = P₁⁰x₁ + P₂⁰x₂
P_T = 200 × 0.312 + 415 × 0.688
Answer : 62.4 + 285.52 = 347.9 mm Hg.

Question 19.
If 1.71 g of sugar (molar mass = 342) is dissolved in 500 cm³ of a solution at 300 K. What will be its osmotic pressure?
Answer:
Data: Mass of solute = W₂ = 1.71 g
Molar mass of solute = M₂ = 342
Volume of solution = 500 cm³ = 0.5 L
Temperature = 300 K
R = 0.083 L bar mol^-1 K^-1

Question 20.
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g water.
Answer:
Data: Vapour pressure of water = P° = 17.535 mm Hg
Mass of solute (glucose) = W₂ = 25 g
Mass of solvent (water) = W₁ = 450 g
Vapour pressure of solution = P = ?
Molecular mass of glucose = (C₆H₁₂O₆)
= 6 × 12 + 12 × 1 + 6 × 16
= 180 g mol^-1
Molecular mass of water = 18 g mol^-1

Question 21.
A 4% solution of a non-volatile solute is isotonic with 0.702% urea solution. Calculate the molar mass of non-volatile solute. (Molar mass of urea 60 g mol^-1)
Answer:
Data:
4% (%) solution of non-volatile solute means 40 g/litre.
0.702% (%) urea solution means. 7.02 g urea/litre.
Molar Mass of urea 60 g mol^-1