JEE Chemistry Solutions MCQs Set 09

Chemistry Objective Questions and Answers: Solutions

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Multiple Choice Questions

Question 1: If a thin slice of sugar beet is placed in a concentrated solution of NaCl, then
(a) sugar beet will lose water from its cells
(b) sugar beet will absorb water from the solution
(c) sugar beet will neither absorb nor lose water
(d) sugar beet will dissolve in the solution
Show Answer & Explanation

Answer: (a) sugar beet will lose water from its cells

Explanation:
1. The salt solution is hypertonic, so water leaves the cells by osmosis (plasmolysis).

Teacher's Note:
1. Water moves from lower to higher solute concentration.

Question 2: In osmosis, the net flow of
(a) solvent molecules occurs from higher concentration to lower concentration
(b) solvent molecules occurs from lower concentration to higher concentration
(c) solute molecules occurs from higher concentration to lower concentration
(d) solute molecules occurs from lower concentration to higher concentration
Show Answer & Explanation

Answer: (b) solvent molecules occurs from lower concentration to higher concentration

Explanation:
1. A semipermeable membrane passes only solvent.
2. Net solvent flow is from the dilute solution to the concentrated solution (where the solvent's concentration is lower).

Teacher's Note:
1. Osmosis tends to equalise concentrations.

Question 3: At constant temperature the osmotic pressure of a solution is
(a) directly proportional to the concentration
(b) inversely proportional to the concentration
(c) directly proportional to the square of the concentration
(d) directly proportional to the square root of the concentration
Show Answer & Explanation

Answer: (a) directly proportional to the concentration

Explanation:
1. π = CRT, so π ∝ C at constant T.

Teacher's Note:
1. van't Hoff's law resembles the ideal gas equation.

Question 4: Which inorganic precipitate acts as a semipermeable membrane?
(a) Calcium sulphate
(b) Barium oxalate
(c) Nickel phosphate
(d) Copper ferrocyanide
Show Answer & Explanation

Answer: (d) Copper ferrocyanide

Explanation:
1. Cu2[Fe(CN)6] deposited in the pores of porcelain is the classic semipermeable membrane.

Teacher's Note:
1. It is used in the Morse-Frazer experiment.

Question 5: If a 0.1 M solution of glucose and a 0.1 M solution of urea are placed on two sides of a semipermeable membrane to equal heights, then it will be correct to say that
(a) there will be no net movement across the membrane
(b) glucose will flow towards the urea solution
(c) urea will flow towards the glucose solution
(d) water will flow from the urea solution towards the glucose solution
Show Answer & Explanation

Answer: (a) there will be no net movement across the membrane

Explanation:
1. Both are non-electrolytes at the same molarity, so they have equal osmotic pressure (isotonic).
2. No net flow occurs.

Teacher's Note:
1. Equal molar concentrations of non-electrolytes are isotonic.

Question 6: The best colligative property used for the determination of molecular masses of polymers is
(a) relative lowering of vapour pressure
(b) osmotic pressure
(c) elevation in boiling point
(d) depression in freezing point
Show Answer & Explanation

Answer: (b) osmotic pressure

Explanation:
1. Polymer solutions are dilute in molarity, so the other effects are too small.
2. Osmotic pressure is large enough to measure even at low molarity and works at room temperature.

Teacher's Note:
1. Osmotic pressure measures large molar masses at room temperature without heating.

Question 7: If the mole fraction of the solvent in a solution decreases, then
(a) the vapour pressure of the solution increases
(b) the boiling point decreases
(c) the osmotic pressure increases
(d) all are correct
Show Answer & Explanation

Answer: (c) the osmotic pressure increases

Explanation:
1. Lower solvent mole fraction means more solute: vapour pressure decreases, boiling point rises, osmotic pressure rises.

Teacher's Note:
1. Only (c) is correct.

Question 8: The osmotic pressure of a solution increases if
(a) the temperature is lowered
(b) the volume is increased
(c) the number of solute molecules is increased
(d) none of these
Show Answer & Explanation

Answer: (c) the number of solute molecules is increased

Explanation:
1. π = (n/V)RT increases with the number of solute molecules, with higher T or with smaller V.

Teacher's Note:
1. Cooling and dilution both decrease π.

Question 9: Which of the following solutions at the same temperature will be isotonic?
(a) 3.42 g of cane sugar in one litre of water and 0.18 g of glucose in one litre of water
(b) 3.42 g of cane sugar in one litre of water and 0.18 g of glucose in 0.1 litre of water
(c) 3.42 g of cane sugar in one litre of water and 0.585 g of NaCl in one litre of water
(d) 3.42 g of cane sugar in one litre of water and 1.17 g of NaCl in one litre of water
Show Answer & Explanation

Answer: (b) 3.42 g of cane sugar in one litre of water and 0.18 g of glucose in 0.1 litre of water

Explanation:
1. Sugar: 3.42/342 = 0.01 M.
2. Glucose: 0.18 g in 0.1 L = 0.001 mol/0.1 L = 0.01 M (isotonic).
3. NaCl solutions have i = 2 giving 0.02 M.

Teacher's Note:
1. Compare the total particle concentration, not mass.

Question 10: If the density of a 2 molal sucrose solution is 1.4 g mL−1 at 25°C, the osmotic pressure is (R = 0.0821 L atm K−1 mol−1)
(a) 4.06 atm
(b) 2 atm
(c) 40.6 atm
(d) 3.4 atm
Show Answer & Explanation

Answer: (c) 40.6 atm

Explanation:
1. Mass of 2 mol sucrose + 1 kg water = 684 + 1000 = 1684 g; volume = 1684/1.4 = 1203 mL.
2. C = 2/1.203 = 1.66 M.
3. π = 1.66 × 0.0821 × 298 = 40.6 atm.

Teacher's Note:
1. Convert molality to molarity using the density.

Question 11: Equal volumes of 0.1 M urea and 0.1 M glucose are mixed. The mixture will have
(a) lower osmotic pressure
(b) the same osmotic pressure
(c) higher osmotic pressure
(d) none of these
Show Answer & Explanation

Answer: (b) the same osmotic pressure

Explanation:
1. The mixture is still 0.1 M in total non-electrolyte particles.
2. Osmotic pressure remains the same.

Teacher's Note:
1. Mixing equimolar non-electrolytes keeps the concentration constant.

Question 12: The osmotic pressure of a solution (density 1 g mL−1) containing 3 g of glucose (molar mass 180) in 60 g of water at 15°C is (R = 0.0821 L atm K−1 mol−1)
(a) 0.34 atm
(b) 0.65 atm
(c) 6.25 atm
(d) 5.57 atm
Show Answer & Explanation

Answer: (c) 6.25 atm

Explanation:
1. Moles = 3/180 = 0.01667; volume of solution ≈ 63 mL = 0.063 L.
2. C = 0.2646 M.
3. π = 0.2646 × 0.0821 × 288 = 6.26 atm.

Teacher's Note:
1. Use the volume of solution (water plus solute), not just of water.

Question 13: The osmotic pressure of a sugar solution at 24°C is 2.5 atmosphere. The concentration of the solution in mole per litre is
(a) 10.25
(b) 1.025
(c) 1025
(d) 0.1025
Show Answer & Explanation

Answer: (d) 0.1025

Explanation:
1. C = π/RT = 2.5/(0.0821 × 297) = 0.1025 mol L−1.

Teacher's Note:
1. Convert 24°C to 297 K.

Question 14: A solution containing 4 g of a non-volatile organic solute per 100 mL was found to have an osmotic pressure equal to 500 cm of mercury at 27°C. The molecular weight of the solute is (R = 0.0821 L atm K−1 mol−1)
(a) 14.97
(b) 149.7
(c) 1697
(d) 1.497
Show Answer & Explanation

Answer: (b) 149.7

Explanation:
1. π = 500 cm Hg = 5000 mm/760 = 6.58 atm.
2. M = wRT/(πV) = 4 × 0.0821 × 300/(6.58 × 0.1) = 149.7 g mol−1.

Teacher's Note:
1. Convert cm of Hg to atm by dividing mm Hg by 760.

Question 15: A solution containing 8.6 g urea in one litre was found to be isotonic with a 0.5% (wt/vol) solution of an organic non-volatile solute. The molecular weight of the latter is
(a) 348.9
(b) 34.89
(c) 3489
(d) 861.2
Show Answer & Explanation

Answer: (b) 34.89

Explanation:
1. Urea: 8.6/60 = 0.1433 M.
2. 0.5% w/v = 5 g L−1, so 5/M = 0.1433 and M = 34.9 g mol−1.

Teacher's Note:
1. Isotonic solutions have equal molar concentration.

Question 16: If a 6.84% (w/V) solution of cane sugar (molar mass 342) is isotonic with a 1.52% (w/V) solution of thiocarbamide, then the molar mass of thiocarbamide is
(a) 152
(b) 76
(c) 60
(d) 180
Show Answer & Explanation

Answer: (b) 76

Explanation:
1. Sugar: 68.4/342 = 0.2 M.
2. Thiocarbamide: 15.2/M = 0.2, so M = 76 g mol−1.

Teacher's Note:
1. Convert w/V percentages to g L−1 by multiplying by 10.

Question 17: A solution containing 500 g of a protein per litre is isotonic with a solution containing 3.42 g of sucrose per litre. The molar mass of the protein is
(a) 5
(b) 146
(c) 34200
(d) 50000
Show Answer & Explanation

Answer: (d) 50000

Explanation:
1. Sucrose: 3.42/342 = 0.01 M.
2. Protein: 500/M = 0.01, so M = 50000 g mol−1.

Teacher's Note:
1. Isotonic means same molar concentration.

Question 18: The osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose that will be isotonic with blood is ____ w/V (R = 0.0821 L atm K−1 mol−1)
(a) 5.41%
(b) 54.1%
(c) 3.5%
(d) 4.53%
Show Answer & Explanation

Answer: (a) 5.41%

Explanation:
1. C = π/RT = 7.65/(0.0821 × 310) = 0.3006 M.
2. Mass per litre = 0.3006 × 180 = 54.1 g, i.e. 5.41 g per 100 mL = 5.41% w/V.

Teacher's Note:
1. Express the answer as grams per 100 mL.

Question 19: A solution containing 10 g per dm3 of urea (molar mass 60 g mol−1) is isotonic with a 5% solution of a non-volatile solute. The molar mass of this non-volatile solute is
(a) 250 g mol−1
(b) 300 g mol−1
(c) 350 g mol−1
(d) 200 g mol−1
Show Answer & Explanation

Answer: (b) 300 g mol−1

Explanation:
1. Urea: 10/60 = 0.1667 M.
2. 5% = 50 g dm−3, so 50/M = 0.1667 and M = 300 g mol−1.

Teacher's Note:
1. Equate molar concentrations.

Question 20: During osmosis, the flow of water through a semipermeable membrane is
(a) from both sides of the semipermeable membrane with equal flow rates
(b) from both sides of the semipermeable membrane with unequal flow rates
(c) from solution having lower concentration only
(d) from solution having higher concentration only
Show Answer & Explanation

Answer: (b) from both sides of the semipermeable membrane with unequal flow rates

Explanation:
1. Water molecules pass in both directions; the net flow towards the concentrated solution is greater because the flow rates are unequal.

Teacher's Note:
1. Net osmosis is the difference of two opposing flows.

Question 21: 200 mL of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of the solution at 300 K is found to be 2.57 × 10−3 bar. The molar mass of the protein will be (R = 0.083 L bar mol−1 K−1)
(a) 61038 g mol−1
(b) 51022 g mol−1
(c) 122044 g mol−1
(d) 31011 g mol−1
Show Answer & Explanation

Answer: (a) 61038 g mol−1

Explanation:
1. M = wRT/(πV) = 1.26 × 0.083 × 300/(2.57 × 10−3 × 0.2) = 61038 g mol−1.

Teacher's Note:
1. Use litres for volume and bar with R in L bar.

Question 22: The osmotic pressure of a solution containing 1 g of protein in 100 mL solution at 300 K is 1.66 bar. The molecular weight of the protein is (R = 0.083 L bar mol−1 K−1)
(a) 150
(b) 120
(c) 130
(d) 300
Show Answer & Explanation

Answer: (a) 150

Explanation:
1. M = 1 × 0.083 × 300/(1.66 × 0.1) = 150 g mol−1.

Teacher's Note:
1. Rearrange π = wRT/(MV).

Question 23: 1 g of a protein polymer of molar mass 16 × 104 is present in 800 mL solution at 300 K. The osmotic pressure in pascal is (R = 8.314 J mol−1 K−1)
(a) 0.4
(b) 0.02
(c) 0.04
(d) 0.2
Show Answer & Explanation

Answer: (b) 0.02

Explanation:
1. n = 1/160000 = 6.25 × 10−6 mol; V = 8 × 10−4 m3.
2. π = nRT/V = 6.25 × 10−6 × 8.314 × 300/(8 × 10−4) = 0.0195 Pa, about 0.02 Pa.

Teacher's Note:
1. Use SI units: V in m3 when R is in J.

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