JEE Chemistry Solutions MCQs Set 13

Chemistry Objective Questions and Answers: Solutions

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Multiple Choice Questions

Question 1: 8 g of NaOH is dissolved in one litre of solution. Its molarity is:
(a) 0.1 M
(b) 0.8 M
(c) 0.4 M
(d) 0.2 M
Show Answer & Explanation

Answer: (d) 0.2 M

Explanation:
1. Moles of NaOH \( = \dfrac{8}{40} = 0.2 \).
2. Molarity \( = \dfrac{0.2}{1\ \text{L}} = 0.2 \) M.

Teacher's Note:
1. Molarity is moles of solute per litre of solution. The molar mass of NaOH is 40 g/mol.

Question 2: 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) water will flow from the urea solution towards the glucose solution
(b) there will be no net movement across the membrane
(c) urea will flow towards the glucose solution
(d) glucose will flow towards the urea solution
Show Answer & Explanation

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

Explanation:
1. Both solutions have the same molarity of non-dissociating solute, so they have the same osmotic pressure (they are isotonic).
2. There is therefore no net flow of solvent in either direction.

Teacher's Note:
1. Osmotic pressure depends only on the number of dissolved particles per litre.

Question 3: A solution containing 10 g per dm³ of urea (molecular mass = 60 g mol⁻¹) is isotonic with a 5% solution of a non-volatile solute. The molecular mass of this non-volatile solute is:
(a) 200 g mol⁻¹
(b) 250 g mol⁻¹
(c) 350 g mol⁻¹
(d) 300 g mol⁻¹
Show Answer & Explanation

Answer: (d) 300 g mol⁻¹

Explanation:
1. A 5% solution has 50 g per litre. Isotonic solutions have equal molarity: \( \dfrac{10}{60} = \dfrac{50}{M} \).
2. \( M = 300 \) g/mol.

Teacher's Note:
1. Isotonic solutions have the same molar concentration, so the molar mass follows from the ratio of mass concentrations.

Question 4: Which of the following solutions has the highest normality?
(a) 0.5 M H2SO4
(b) 6 g of NaOH per 100 mL
(c) 8 g of KOH per litre
(d) 1 N phosphoric acid
Show Answer & Explanation

Answer: (b) 6 g of NaOH per 100 mL

Explanation:
1. KOH: \( 8/56 = 0.14 \) N. Phosphoric acid: 1 N.
2. NaOH: 60 g per litre gives \( 60/40 = 1.5 \) N. H2SO4: \( 0.5\times2 = 1 \) N.
3. The highest is 1.5 N.

Teacher's Note:
1. Convert each to normality before comparing. The mass per 100 mL must be scaled to a litre.

Question 5: If 18 g of glucose is present in 1000 g of solvent, the solution is said to be:
(a) 0.1 molal
(b) 0.1 molar
(c) 1 molar
(d) 0.5 molar
Show Answer & Explanation

Answer: (a) 0.1 molal

Explanation:
1. Moles of glucose \( = \dfrac{18}{180} = 0.1 \).
2. The amount of solvent is 1000 g, so the concentration is 0.1 mol per kg of solvent = 0.1 molal.

Teacher's Note:
1. Per kilogram of solvent means molal. Per litre of solution means molar.

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

Answer: (d) osmotic pressure

Explanation:
1. Polymers and proteins have very large molar masses, so their solutions of measurable mass concentration are extremely dilute in moles.
2. Osmotic pressure gives a measurable value even at such low molarity and is measured at room temperature, which avoids damage to the macromolecule.

Teacher's Note:
1. The other colligative properties give changes that are too small to measure for large molecules.

Question 7: 1.00 g of a non-electrolyte solute (molar mass 250 g mol⁻¹) was dissolved in 51.2 g of benzene. If the freezing point depression constant Kf of benzene is 5.12 K kg mol⁻¹, the freezing point of benzene will be lowered by:
(a) 0.3 K
(b) 0.2 K
(c) 0.4 K
(d) 0.5 K
Show Answer & Explanation

Answer: (c) 0.4 K

Explanation:
1. \( m = \dfrac{1/250}{0.0512} = 0.078 \) mol/kg.
2. \( \Delta T_f = 5.12\times0.078 = 0.4 \) K.

Teacher's Note:
1. Convert the solvent mass to kilograms before calculating the molality.

Question 8: An aqueous solution of glucose is 10% in strength (w/V). The volume in which 2 g mole of it is dissolved will be:
(a) 18 litre
(b) 1.8 litre
(c) 3.6 litre
(d) 0.9 litre
Show Answer & Explanation

Answer: (c) 3.6 litre

Explanation:
1. 2 mol of glucose \( = 360 \) g. A 10% solution has 100 g in 1 L.
2. \( V = \dfrac{360}{100} = 3.6 \) L.

Teacher's Note:
1. 10% w/V means 10 g in 100 mL, i.e. 100 g per litre.

Question 9: The molarity of a solution of sodium chloride (molar mass = 58.5 g/mol) in water containing 5.85 g of sodium chloride in 500 mL of solution is:
(a) 0.25
(b) 1.0
(c) 0.2
(d) 2.0
Show Answer & Explanation

Answer: (c) 0.2

Explanation:
1. Moles \( = \dfrac{5.85}{58.5} = 0.1 \).
2. Molarity \( = \dfrac{0.1}{0.5} = 0.2 \) M.

Teacher's Note:
1. Convert 500 mL to 0.5 L before dividing.

Question 10: If the mole fraction of the solvent in a solution decreases, then:
(a) the vapour pressure of the solution increases
(b) the osmotic pressure increases
(c) all are correct
(d) the boiling point decreases
Show Answer & Explanation

Answer: (b) the osmotic pressure increases

Explanation:
1. A lower mole fraction of the solvent means more solute is present.
2. The vapour pressure decreases and the boiling point increases, but the osmotic pressure (proportional to the solute concentration) increases.

Teacher's Note:
1. Statements (1) and (2) are the opposite of what actually happens.

Question 11: A solution of acetone in ethanol:
(a) shows a positive deviation from Raoult's law
(b) shows a negative deviation from Raoult's law
(c) behaves like a near ideal solution
(d) obeys Raoult's law
Show Answer & Explanation

Answer: (a) shows a positive deviation from Raoult's law

Explanation:
1. Ethanol molecules are hydrogen-bonded to each other. Adding acetone breaks some of these bonds, and the acetone-ethanol interactions are weaker.
2. Weaker A-B interactions give a higher vapour pressure than predicted, which is a positive deviation.

Teacher's Note:
1. Breaking of hydrogen bonds on mixing leads to positive deviation.

Question 12: One litre of N/2 HCl solution is heated in a beaker. It was observed that when the volume of the solution was reduced to 600 mL, 3.25 g of HCl was lost. Calculate the normality of the new solution:
(a) 0.171
(b) 0.342
(c) 0.685
(d) 1.70
Show Answer & Explanation

Answer: (c) 0.685

Explanation:
1. Initially \( 0.5\times36.5 = 18.25 \) g of HCl. After the loss: \( 18.25-3.25 = 15 \) g.
2. \( N = \dfrac{15/36.5}{0.6} = 0.685 \) N.

Teacher's Note:
1. Subtract the HCl lost from the initial amount, then divide by the new volume.

Question 13: For preparing a 0.1 M solution of H2SO4 in one litre, we need H2SO4:
(a) 4.9 g
(b) 0.98 g
(c) 49.0 g
(d) 9.8 g
Show Answer & Explanation

Answer: (d) 9.8 g

Explanation:
1. Moles needed \( = 0.1\times1 = 0.1 \).
2. Mass \( = 0.1\times98 = 9.8 \) g.

Teacher's Note:
1. Mass = molarity × volume (L) × molar mass.

Question 14: 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. \( \pi = \dfrac{n}{V}RT \).
2. It increases with higher n (more solute particles), with higher T, or with lower V.

Teacher's Note:
1. Osmotic pressure is proportional to concentration and to the absolute temperature.

Question 15: During osmosis, the flow of water through a semipermeable membrane is:
(a) from the solution having lower concentration only
(b) from both sides of the semipermeable membrane with equal flow rates
(c) from the solution having higher concentration only
(d) from both sides of the semipermeable membrane with unequal flow rates
Show Answer & Explanation

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

Explanation:
1. Water molecules pass through the membrane in both directions.
2. The rate from the dilute side is greater, so there is a net flow towards the concentrated side.

Teacher's Note:
1. Osmosis is a net flow. Molecules still cross in both directions.

Question 16: The mole fraction of the solute in a one molal aqueous solution is:
(a) 0.027
(b) 0.009
(c) 0.018
(d) 0.036
Show Answer & Explanation

Answer: (c) 0.018

Explanation:
1. \( x = \dfrac{1}{1+55.5} = 0.0177\approx0.018 \).

Teacher's Note:
1. 1000 g of water is 55.5 mol.

Question 17: The mole fraction of glycerine (C3H5(OH)3) in a solution of 36 g of water and 46 g of glycerine is:
(a) 0.46
(b) 0.40
(c) 0.36
(d) 0.20
Show Answer & Explanation

Answer: (d) 0.20

Explanation:
1. Moles of water \( = 36/18 = 2 \). Moles of glycerine \( = 46/92 = 0.5 \).
2. \( x = \dfrac{0.5}{2+0.5} = 0.20 \).

Teacher's Note:
1. The molar mass of glycerine is 92 g/mol (C3H8O3).

Question 18: Which of the following solutions at the same temperature will be isotonic? (Molar masses: cane sugar 342, glucose 180, NaCl 58.5)
(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 1.17 g of NaCl in one litre of water
(d) 3.42 g of cane sugar in one litre of water and 0.585 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. Cane sugar: \( 3.42/342 = 0.01 \) mol in 1 L = 0.01 M.
2. Option 2: glucose \( 0.18/180 = 0.001 \) mol in 0.1 L = 0.01 M. The concentrations are equal.
3. The NaCl solutions in options 3 and 4 give 0.02 M and 0.04 M of particles, and option 1 gives 0.001 M.

Teacher's Note:
1. Isotonic solutions have equal total particle concentrations.

Question 19: A 5% solution (by mass) of cane sugar in water has a freezing point of 271 K and the freezing point of pure water is 273.15 K. The freezing point of a 5% solution (by mass) of glucose in water is:
(a) 269.07 K
(b) 273.15 K
(c) 271 K
(d) 277.23 K
Show Answer & Explanation

Answer: (a) 269.07 K

Explanation:
1. For cane sugar: \( \Delta T_f = 2.15 \) K, proportional to \( \dfrac{5}{342} \). For glucose the same mass gives \( \dfrac{5}{180} \) moles.
2. \( \Delta T_f(glucose) = 2.15\times\dfrac{342}{180} = 4.08 \) K.
3. Freezing point \( = 273.15-4.08 = 269.07 \) K.

Teacher's Note:
1. For equal mass percent, the smaller molar mass gives more moles and a bigger depression.

Question 20: The density of a 2.05 M solution of acetic acid in water is 1.02 g mL⁻¹. The molality of the solution is:
(a) 0.44 mol kg⁻¹
(b) 1.14 mol kg⁻¹
(c) 2.28 mol kg⁻¹
(d) 3.28 mol kg⁻¹
Show Answer & Explanation

Answer: (c) 2.28 mol kg⁻¹

Explanation:
1. 1 L of solution has mass 1020 g and contains \( 2.05\times60 = 123 \) g of acetic acid.
2. Water \( = 1020-123 = 897 \) g, so \( m = \dfrac{2.05}{0.897} = 2.28 \) mol/kg.

Teacher's Note:
1. To convert molarity to molality, find the mass of the solvent in 1 L of solution.

Question 21: 1000 g of an aqueous solution of CaCO3 contains 10 g of calcium carbonate. The concentration of the solution is:
(a) 10 ppm
(b) 10,000 ppm
(c) 1000 ppm
(d) 100 ppm
Show Answer & Explanation

Answer: (b) 10,000 ppm

Explanation:
1. \( \text{ppm} = \dfrac{10}{1000}\times10^6 = 10^4 \).

Teacher's Note:
1. 1% by mass is equal to 10,000 ppm.

Question 22: If the density of a 2 molal sucrose solution is 1.4 g mL⁻¹ at 25 °C, find its osmotic pressure.
(a) 40.6 atm
(b) 3.4 atm
(c) 4.06 atm
(d) 2 atm
Show Answer & Explanation

Answer: (a) 40.6 atm

Explanation:
1. Mass of solution containing 1 kg of water \( = 1000+2\times342 = 1684 \) g, volume \( = 1684/1.4 = 1203 \) mL.
2. Molarity \( = \dfrac{2}{1.203} = 1.66 \) M.
3. \( \pi = CRT = 1.66\times0.0821\times298 \approx 40.6 \) atm.

Teacher's Note:
1. Convert molality to molarity using the density of the solution, then apply π = CRT.

Question 23: A 0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol⁻¹, the lowering in the freezing point of the solution is:
(a) -1.12 K
(b) 0.56 K
(c) 1.12 K
(d) -0.56 K
Show Answer & Explanation

Answer: (c) 1.12 K

Explanation:
1. \( i = 1+\alpha = 1.2 \) for a monobasic acid.
2. \( \Delta T_f = iK_fm = 1.2\times1.86\times0.5 = 1.12 \) K.
3. The lowering is 1.12 K (the freezing point itself falls to -1.12 °C).

Teacher's Note:
1. The lowering is a positive quantity. A negative sign belongs to the new freezing point, not to the depression.

Question 24: What will be the density (in g mL⁻¹) of a 3.60 molar sulphuric acid solution having 29% H2SO4 by mass? (Molar mass = 98 g mol⁻¹)
(a) 1.22
(b) 1.45
(c) 1.88
(d) 1.64
Show Answer & Explanation

Answer: (a) 1.22

Explanation:
1. \( M = \dfrac{\%\times d\times10}{98} \Rightarrow d = \dfrac{3.6\times98}{29\times10} = 1.217 \).

Teacher's Note:
1. Use M = (percent × density × 10) / molar mass and rearrange for density.

Question 25: What is the normality of a 1 M H3PO4 solution?
(a) 0.5 N
(b) 2.0 N
(c) 3.0 N
(d) 1.0 N
Show Answer & Explanation

Answer: (c) 3.0 N

Explanation:
1. H3PO4 is tribasic, so its n-factor is 3.
2. \( N = M\times n = 1\times3 = 3 \) N.

Teacher's Note:
1. Normality = molarity × n-factor (basicity for acids, acidity for bases).

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