Question 1

Given below are two statements: Given: Molar mass of C, H, O, Cl are 12, 1, 16 and 35.5 g mol^ -1 , respectively. Statement I: In 30\% (w/w) solution of methanol in CCl_4 (at T K), the mole fraction o

Accepted Answer

Statement I: 30\% (w/w) solution of methanol in CCl_4 means 30 g of methanol is present in 70 g of CCl_4. Molar mass of methanol (CH_3OH) = 12 + 4 1 + 16 = 32 g mol^ -1 Molar mass of CCl_4 = 12 + 4 35.5 = 154 g mol^ -1 Moles of methanol = 30 32 = 0.9375 mol Moles of CCl_4 = 70 154 = 0.4545 mol Mole fraction of CCl_4 = 0.4545 0.9375 + 0.4545 = 0.4545 1.392 0.326 0.33 Thus, Statement I is true. Statement II: In pure methanol, molecules are held together by strong intermolecular hydrogen bonding. O

Question 2

Which of the following statements are not correct? A. For water, magnitude of K_b is more than the magnitude of K_f. B. The elevation in boiling point of water when a non-volatile solute is added to i

Accepted Answer

For water, the molal elevation constant K_b = 0.52 K kg mol ^ -1 and the molal depression constant K_f = 1.86 K kg mol ^ -1 . Thus, K_b For a solution with a given molality m of a non-volatile solute, the elevation in boiling point is T_b = K_b m and the depression in freezing point is T_f = K_f m. Since K_b Osmotic pressure measurement is preferred for determining the molar mass of macromolecules such as proteins and polymers because the measurement is done at room temperature and the magnitude

Question 3

Given below are two statements : Statement I : H_2O molecules move from the chamber 1 to chamber 2. Statement II : The osmotic pressure of a solution prepared by dissolving 50 mg of potassium sulphate

Accepted Answer

For Statement I: Molarity of the solution in Chamber 1 (M_1) = 18 / 180 100 / 1000 = 1 M Molarity of the solution in Chamber 2 (M_2) = 30 / 180 250 / 1000 = 1/6 1/4 = 0.667 M Since M_1 > M_2, the solution in Chamber 1 has a higher solute concentration. During osmosis, the solvent (water) moves from the region of lower solute concentration to the region of higher solute concentration across a semi-permeable membrane. Thus, water will move from Chamber 2 to Chamber 1. Statement I is false. For Sta

Question 4

When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a solvent the vapor pressure of solution was x\% of vapor pressure of pure solvent. What is x\%?

Accepted Answer

Moles of solute, n_B = 0.25 Moles of solvent, n_A = 1 Mole fraction of solvent, X_A = n_A n_A + n_B = 1 1 + 0.25 = 1 1.25 = 0.8 According to Raoult's law, the vapor pressure of the solution is given by P_s = P^0 X_A P_s = 0.8 P^0 = 80\% of P^0 Thus, the vapor pressure of the solution is 80\% of the vapor pressure of the pure solvent. Answer: 80\%

Question 5

20 g hemoglobin in a 1 L aqueous solution (A) at 300 K is separated from pure water by semi permeable membrane. At equilibrium the height of solution in a tube dipped in a solution (A) is found to be

Accepted Answer

The osmotic pressure of the solution is balanced by the hydrostatic pressure of the liquid column. = g h Given: Density of solution, = 1000 kg m ^ -3 Acceleration due to gravity, g = 10 m s ^ -2 Height difference, h = 80.0 mm = 0.08 m = 1000 10 0.08 = 800 Pa = 0.8 kPa From the van 't Hoff equation for osmotic pressure: = C R T = n V R T Given: Volume of solution, V = 1 L = 1 dm ^3 Gas constant, R = 8.3 kPa dm ^3 K ^ -1 mol ^ -1 Temperature, T = 300 K Substituting the values: 0.8 = n 1 8.3 300 n

Question 6

What is the mole fraction of water in 10% by weight (w/w) of aqueous urea solution? [Given: Molar mass of H, O, C and N are 1, 16, 12 and 14 g mol^ -1 respectively.]

Accepted Answer

10\% w/w aqueous urea solution means 10 g of urea is present in 100 g of solution. Mass of urea = 10 g Mass of water = 100 - 10 = 90 g Molar mass of urea ( NH _2 CONH _2) = 14 + 2 + 12 + 16 + 14 + 2 = 60 g mol^ -1 Molar mass of water ( H _2 O ) = 18 g mol^ -1 Moles of urea, n_ urea = 10 60 = 1 6 mol Moles of water, n_ water = 90 18 = 5 mol Total moles = 5 + 1 6 = 31 6 mol Mole fraction of water, x_ water = n_ water n_ water + n_ urea = 5 31 6 = 30 31 0.967 Answer: 0.967

Question 7

At 27°C, 0.1 M, 1 L K_4[Fe(CN)_6] aqueous solution and 0.1 M, 1 L FeCl_3 aqueous solution are placed in a container separated by a semi permeable membrane AB. Assume complete dissociation of both the

Accepted Answer

For complete dissociation, the van 't Hoff factor i is equal to the number of ions produced per formula unit. For side 'x' containing K_4[Fe(CN)_6]: K_4[Fe(CN)_6] 4K^+ + [Fe(CN)_6]^ 4- Number of ions, i_x = 5 Effective concentration (osmolarity) = i_x C = 5 0.1 = 0.5 M For side 'y' containing FeCl_3: FeCl_3 Fe^ 3+ + 3Cl^- Number of ions, i_y = 4 Effective concentration (osmolarity) = i_y C = 4 0.1 = 0.4 M Since the effective concentration of side 'y' (0.4 M) is less than that of side 'x' (0.5 M)

Question 8

A non-volatile, non-electrolyte solid solute when dissolved in 40 g of a solvent, the vapour pressure of the solvent decreased from 760 mm Hg to 750 mm Hg. If the same solution boils at 320 K, then th

Accepted Answer

From the relative lowering of vapour pressure, we have: P^0 - P_s P_s = n_B n_A where n_B is the number of moles of solute and n_A is the number of moles of solvent. Substituting the given values: 760 - 750 750 = n_B n_A 10 750 = n_B n_A n_A = 75 n_B The elevation in boiling point is given by: T_b = T_b - T_b^0 = 320 - 319.5 = 0.5 K We also know that T_b = K_b m, where m is the molality of the solution. m = n_B Mass of solvent in kg = n_B 0.04 Substituting the values into the elevation in boilin

Question 9

Solution A is prepared by dissolving 1 g of a protein (molar mass = 50000 g mol^ -1 ) in 0.5 L of water at 300 K. Its osmotic pressure is x bar. Solution B is made by dissolving 2 g of same protein in

Accepted Answer

Osmotic pressure is given by = CRT = w MV RT For solution A: x = 1 50000 0.5 0.083 300 x = 1 25000 24.9 = 9.96 10^ -4 bar For solution B: y = 2 50000 1 0.083 300 y = 2 50000 24.9 = 9.96 10^ -4 bar For the resultant solution (mixture of A and B): Total mass of protein = 1 + 2 = 3 g Total volume of solution = 0.5 + 1 = 1.5 L z = 3 50000 1.5 0.083 300 z = 3 75000 24.9 = 9.96 10^ -4 bar Therefore, x = y = z = 9.96 10^ -4 bar. Answer: 9.96 10^ -4 ; 9.96 10^ -4 ; 9.96 10^ -4

Question 10

19.5 g of fluoro acetic acid (molar mass = 78 g mol^ -1 ) is dissolved in 500 g of water at 298 K. The depression in the freezing point of water was 1°C. What is K_a of fluoro acetic acid? (For water,

Accepted Answer

The molality of the solution is given by: m = Mass of solute Molar mass of solute Mass of solvent in kg m = 19.5 78 0.5 = 0.5 mol kg ^ -1 The depression in freezing point is given by: T_f = i K_f m Substituting the given values: 1 = i 1.86 0.5 1 = i 0.93 i = 1 0.93 = 100 93 For the dissociation of a weak acid HA H^+ + A^-, the van 't Hoff factor is i = 1 + , where is the degree of dissociation. 1 + = 100 93 = 100 93 - 1 = 7 93 The acid dissociation constant K_a is given by: K_a = C ^2 1 - Since

Question 11

Given below are two statements : Statement (I) : 1,2,3-Trihydroxypropane can be separated from water by simple distillation. Statement (II) : An azeotropic mixture cannot be separated by fractional di

Accepted Answer

Statement I: 1,2,3-Trihydroxypropane (glycerol) has a very high boiling point (around 290^ C ) and it decomposes at its boiling point under normal atmospheric pressure. Therefore, it cannot be separated from water by simple distillation. Instead, it is separated by distillation under reduced pressure (vacuum distillation). Hence, Statement I is false. Statement II: An azeotropic mixture is a constant boiling mixture in which the liquid and vapour phases have the exact same composition. Because t

Question 12

Consider the following aqueous solutions. I. 2.2 g Glucose in 125 mL of solution. II. 1.9 g Calcium chloride in 250 mL of solution. III. 9.0 g Urea in 500 mL of solution. IV. 20.5 g Aluminium sulphate

Accepted Answer

T_b = i k_b m For dilute solution (M = m) Molarity i m (I) M_ glucose = 2.2 180 1000 125 = 0.098 0.098 1 (II) M_ CaCl_2 = 1.9 111 1000 250 = 0.068 0.068 3 (III) M_ urea = 9 60 1000 500 = 0.3 0.3 1 (IV) M_ Al_2(SO_4)_2 = 20.5 342 1000 750 0.08 0.08 5 Order of T_b = Al_2(SO_4)_3 > Urea > CaCl_2 > Glucose So order of BP = Al_2(SO_4)_3 > Urea > CaCl_2 > Glucose So Answer will be I < II < III < IV

Question 13

At T ( K ), 2 moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of ideal solution formed is 320 mm Hg. At this stage, one mole of A and one mole of B are added to the solution.

Accepted Answer

For an ideal solution obeying Raoult's Law, total pressure is P = P_A^o x_A + P_B^o x_B. Initially with 2 moles A and 3 moles B: x_A = 0.4, x_B = 0.6, and P = 320 = 0.4P_A^o + 0.6P_B^o After adding 1 mole each: 3 moles A and 4 moles B gives x_A = 3 7 , x_B = 4 7 , and P = 328.6 = 3 7 P_A^o + 4 7 P_B^o From the first equation: 3200 = 4P_A^o + 6P_B^o From the second: 2300.2 = 3P_A^o + 4P_B^o Solving: We find P_B^o = 200 mm Hg and P_A^o = 500 mm Hg.

Question 14

At 298 K, the mole percentage of N _ 2 ( ~g ) in air is 80 \%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of N _ 2 ( ~g ) in water at 298 K ? ( K _ H for N _ 2

Accepted Answer

Using Henry's Law: Mole fraction = P K_H N₂ mole percentage in air = 80%, total pressure = 10 atm Partial pressure of N₂ = 0.80 × 10 = 8 atm = 8 × 760 = 6080 mmHg K_H for N₂ = 6.5 × 10⁷ mmHg Mole fraction of N₂ = 6080 6.5 10^7 = 6.08 10^3 6.5 10^7 = 0.935 10^ -4 = 9.35 10^ -5

Question 15

Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and 15 kN m ^ -2 respectively. What is the mole fraction of A in solution of A and B

Accepted Answer

For an ideal solution with Raoult's law, the partial pressure of A in vapour is P_A = P_A^0 y_A where y_A = 0.8 is the mole fraction in vapour phase. Thus P_A = 55 0.8 = 44 kNm^ -2 . Using the equilibrium condition y_A = P_A^0 _A P_ total where _A is mole fraction in liquid and P_ total = P_A^0 _A + P_B^0(1- _A): 0.8 = 55 _A 55 _A + 15(1- _A) 0.8(55 _A + 15 - 15 _A) = 55 _A 44 _A + 12 - 12 _A = 55 _A 12 = 23 _A _A = 0.5217

Question 16

A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute 'A' of molar mass 60 ~g ~mol ^ -1 and 0.9 g of a non-volatile non-electrolyte solute ' B ' of molar mass 180 ~g ~mol

Accepted Answer

Calculate osmotic pressure using = nRT V where n is total moles of solute particles. Moles of solute A: n_A = 0.3 g 60 g/mol = 0.005 mol Moles of solute B: n_B = 0.9 g 180 g/mol = 0.005 mol Total moles: n_ total = 0.005 + 0.005 = 0.01 mol Volume of solution: 100 mL = 0.1 L Temperature: 27°C = 300 K Osmotic pressure: = 0.01 0.082 300 0.1 = 0.246 0.1 = 2.46 atm

Question 17

' W ' g of a non-volatile electrolyte solid solute of molar mass ' M ' g mol ^ -1 when dissolved in 100 mL water, decreases vapour pressure of water from 640 mm Hg to 600 mm Hg. If aqueous solution of

Accepted Answer

Given the vapour pressure of pure water P^o = 640 mm Hg and vapour pressure of solution P_s = 600 mm Hg. According to Raoult's law for an electrolyte solution, the relative lowering of vapour pressure is given by P^o - P_s P^o = i x_2, where i is the van't Hoff factor and x_2 is the mole fraction of the solute. Substituting the values: 640 - 600 640 = i x_2 40 640 = i x_2 i x_2 = 1 16 ... (1) The elevation in boiling point is given by T_b = i K_b m, where m is the molality. Given T_b = 375 K and

Question 18

Two liquids A and B form an ideal solution. At 320 K, the vapour pressure of the solution, containing 3 mol of A and 1 mol of B is 500 mm Hg. At the same temperature, if 1 mol of A is further added to

Accepted Answer

For an ideal solution, Raoult's law states: P_ total = P_A^o x_A + P_B^o x_B. Initial condition: 3 mol of A and 1 mol of B give a total vapor pressure of 500 mm Hg. Mole fractions are x_A = 3 4 = 0.75 and x_B = 1 4 = 0.25. This gives: 500 = 0.75 P_A^o + 0.25 P_B^o ... (1). After adding 1 mol of A: 4 mol of A and 1 mol of B give a total vapor pressure of 520 mm Hg. Mole fractions are x_A = 4 5 = 0.8 and x_B = 1 5 = 0.2. This gives: 520 = 0.8 P_A^o + 0.2 P_B^o ... (2). From equation (1) multiplied

Question 19

Which one of the following graphs accurately represents the plot of partial pressure of CS _ 2 vs its mole fraction in a mixture of acetone and CS _ 2 at constant temperature?

Accepted Answer

The partial pressure of a component in a binary mixture follows Raoult's law for ideal solutions: P_ i = x_i P°_i, which is linear. However, the acetone-CS₂ system exhibits positive deviation from ideality due to different molecular interactions. When acetone and CS₂ mix, the hydrogen bonding between acetone molecules is disrupted, and the weaker interactions between unlike molecules allow molecules to escape more easily into the vapor phase. This results in partial pressures higher than predict

Question 20

At T ( K ), 100 ~g of 98 \% H _ 2 SO _ 4 ( w / w ) aqueous solution is mixed with 100 g of 49 \% H _ 2 SO _ 4 ( w / w ) aqueous solution. What is the mole fraction of H _ 2 SO _ 4 in the resultant sol

Accepted Answer

Molar mass of H₂SO₄ = 98 g/mol. Solution 1: 100 g at 98% gives 98 g H₂SO₄ = 1 mol; water = 2 g = 0.111 mol. Solution 2: 100 g at 49% gives 49 g H₂SO₄ = 0.5 mol; water = 51 g = 2.833 mol. Mixed solution: Total H₂SO₄ = 1 + 0.5 = 1.5 mol. Total water = 0.111 + 2.833 = 2.944 mol. Mole fraction of H₂SO₄ = 1.5/(1.5 + 2.944) = 1.5/4.444 = 0.337.

Solutions — Previous Year Questions & Solutions