When 45 ~g of solute is dissolved in 600 ~g water, freezing point gets lowered by 2.2 ~K , calculate molar mass of solute. (K_f=1.86 ~K ~kg ~mol^ -1 )

Correct answer: A. 63.4 ~g ~mol^ -1 aligned & w_1=600 g, Δ T_f=2.2 K & w_2=45 g, K_f=1.86 K kg mol^ -1 & M=(K_f × w_2)/(Δ T_f × w_1)=(1.86 × 45 × 1000)/(2.2 × 600)=63.4 g mol^ -1 aligned

AIIMS 201826 May

Exam: AIIMS
Year: 2018
Date: 26 May
Subject: Chemistry
Chapter: Solutions

AIIMS 2018 · 26 May · Chemistry · Solutions

When 45 ~g of solute is dissolved in 600 ~g water, freezing point...

When

45 \mathrm{~g}

of solute is dissolved in

600 \mathrm{~g}

water, freezing point gets lowered by

2.2 \mathrm{~K}

, calculate molar mass of solute.

\left(K_f=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)

Options

A $63.4 \mathrm{~g} \mathrm{~mol}^{-1}$
B $80 \mathrm{~g} \mathrm{~mol}^{-1}$
C $90 \mathrm{~g} \mathrm{~mol}^{-1}$
D $21 \mathrm{~g} \mathrm{~mol}^{-1}$

Correct answer

A. $63.4 \mathrm{~g} \mathrm{~mol}^{-1}$

Step-by-step solution

\begin{aligned} & \text{ } w_1=600 \mathrm{ g}, \quad \Delta T_f=2.2 \mathrm{ K} \\ & w_2=45 \mathrm{ g}, \quad K_f=1.86 \mathrm{ K} \mathrm{ kg} \mathrm{ mol}^{-1} \\ & M=\frac{K_f \times w_2}{\Delta T_f \times w_1}=\frac{1.86 \times 45 \times 1000}{2.2 \times 600}=63.4 \mathrm{ g} \mathrm{ mol}^{-1}\end{aligned}

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