On dissolving 0.5g of a non-volatile non-ionic solute to 39g of benzene, its vapor pressure decreases from 650mmHg to 640mmHg . The depression of freezing point of benzene (in K ) upon addition of the solute is _______

Correct answer: 0 Δ P Ps = n _ solute n _ solvent (650 - 640)/(640) = 1 × 0.5 × 78 M × 39 M _ solute = 64g Δ T _ F = k _ f × m ⇒ 5.12 × (0.5 × 1000)/(64 × 39) Δ T _ f = 1.0256

JEE Advanced 2019Paper 1

Exam: JEE Advanced
Year: 2019
Paper: Paper 1
Subject: Chemistry
Chapter: Solutions

JEE Advanced 2019 · Paper 1 · Chemistry · Solutions

On dissolving 0.5g of a non-volatile non-ionic solute to 39g of...

On dissolving

0.5g

of a non-volatile non-ionic solute to

39g

of benzene, its vapor pressure decreases from

650mmHg

640mmHg .

The depression of freezing point of benzene (in

K

) upon addition of the solute is _______
(Given data: Molar mass and the molal freezing point depression constant of benzene are

78gmo{l}^{- 1}

and

5.12Kkgmo{l}^{- 1},

respectively)

Correct answer

Step-by-step solution

\dfrac{\text{\Delta P}}{\text{Ps}} = \dfrac{{n}_{\text{solute}}}{{\text{n}}_{\text{solvent}}}

\dfrac{650 - 640}{640} = \dfrac{1 \times 0.5 \times 78}{\text{M} \times 39}

{\text{M}}_{\text{solute}} = 64\text{g}

{\text{\Delta T}}_{\text{F}} = {\text{k}}_{\text{f}} \times \text{m} ⇒

5.12 \times \dfrac{0.5 \times 1000}{64 \times 39}

{\text{\Delta T}}_{\text{f}} = 1.0256

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