Q16. Vapour pressure of an aqueous solution of non-
volatile solute which boils at 101.56°C at 25°C

Q: What is the correct answer for CUET UG Chemistry Q16?

The correct answer is b. 1. Calculate elevation in boiling point (ΔT_b): ΔT_b = 101.56 °C - 100 °C (normal boiling point of water) = 1.56 °C.2. Calculate molality (m) using ΔT_b = K_b × m: 1.56 = 0.52 × m ⇒ m = 1.56 / 0.52 = 3 mol/kg.3. Calculate mole fraction of solute (X_solute): X_solute = m / (m + (1000/M_water)) = 3 / (3 + (1000/18)) = 3 / (3 + 55.56) = 3 / 58.56 ≈ 0.0512.4. Calculate vapor pressure of the solution (

Q: Q16. Vapour pressure of an aqueous solution of non- volatile solute which boils at 101.56°C at 25°C will be (V.P. of pure water at 25°C is 200 mmHg and Kb for water is 0.52 kg mol⁻¹) (a) 198.98…

The correct answer is b. 1. Calculate elevation in boiling point (ΔT_b): ΔT_b = 101.56 °C - 100 °C (normal boiling point of water) = 1.56 °C.2. Calculate molality (m) using ΔT_b = K_b × m: 1.56 = 0.52 × m ⇒ m = 1.56 / 0.52 = 3 mol/kg.3. Calculate mole fraction of solute (X_solute): X_solute = m / (m + (1000/M_water)) = 3 / (3 + (1000/18)) = 3 / (3 + 55.56) = 3 / 58.56 ≈ 0.0512.4. Calculate vapor pressure of the solution (

Q: What are the options for this CUET UG Chemistry question?

A. 198.98 mmHg · B. 189.8 mmHg · C. 179.6 mmHg · D. 184.7 mmHg

Question

Q16. Vapour pressure of an aqueous solution of non-
volatile solute which boils at 101.56°C at 25°C will be
(V.P. of pure water at 25°C is 200 mmHg and Kb for
water is 0.52 kg mol⁻¹)
(a) 198.98 mmHg
(b) 189.8 mmHg
(c) 179.6 mmHg
(d) 184.7 mmHg

FirstInTest · Accepted Answer

1. Calculate elevation in boiling point (ΔT_b): ΔT_b = 101.56 °C - 100 °C (normal boiling point of water) = 1.56 °C.2. Calculate molality (m) using ΔT_b = K_b × m: 1.56 = 0.52 × m ⇒ m = 1.56 / 0.52 = 3 mol/kg.3. Calculate mole fraction of solute (X_solute): X_solute = m / (m + (1000/M_water)) = 3 / (3 + (1000/18)) = 3 / (3 + 55.56) = 3 / 58.56 ≈ 0.0512.4. Calculate vapor pressure of the solution (P_solution) using Raoult's Law: P_solution = P°_water × X_solvent = P°_water × (1 - X_solute). P_solution = 200 mmHg × (1 - 0.0512) = 200 mmHg × 0.9488 = 189.76 mmHg ≈ 189.8 mmHg.

Answer

198.98 mmHg

Answer

189.8 mmHg

Answer

179.6 mmHg

Answer

184.7 mmHg

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