1.80 g of solute A was dissolved in $62.5 \mathrm{~cm}^{3}$ of ethanol and freezing point of the solution was found to be 155.1 K. The molar mass of solute A is ______ $\mathrm{g} \mathrm{~mol}^{-1}$. [Given: Freezing point of ethanol is 156.0 K. Density of ethanol is $0.80 \mathrm{~g} \mathrm{~cm}^{-3}$. Freezing point depression constant of ethanol is $2.00 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$]

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Solutions

1.80 g of solute A was dissolved in

62.5 cm^{3}

of ethanol and freezing point of the solution was found to be 155.1 K. The molar mass of solute A is ______

g mol^{- 1}

.
[Given: Freezing point of ethanol is 156.0 K.
Density of ethanol is

0.80 g cm^{- 3}

.
Freezing point depression constant of ethanol is

2.00 K kg mol^{- 1}

]

Answer: 80

Solution

Mass of ethanol = volume × density =

62.5 \times 0.80 = 50

g = 0.050 kg

Freezing point depression:

Δ T_{f} = 156.0 - 155.1 = 0.9

K

Using

Δ T_{f} = K_{f} \times m

, where

m

is molality:

0.9 = 2.00 \times \frac{1.80 \times 1000}{M _{w} \times 50}

0.9 = \frac{3600}{50 M _{w}}

M_{w} = \frac{3600}{0.9 \times 50} = \frac{3600}{45} = 80

g mol

^{- 1}

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Integer answer

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