83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is $\_\_\_\_$ K. (Nearest integer). [Use: Molal Freezing point depression constant of water $=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ Freezing point of water $=273 \mathrm{~K}$ Atomic masses: C : $12.0 \mathrm{u}, \mathrm{O}: 16.0 \mathrm{u}, \mathrm{H}: 1.0 \mathrm{u}$]

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Solutions

83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is

____

K. (Nearest integer).
[Use: Molal Freezing point depression constant of water

= 1.86 K kg mol^{- 1}

Freezing point of water

= 273 K

Atomic masses: C :

12.0 u, O : 16.0 u, H : 1.0 u

]

Answer: 269

Solution

Ethylene glycol is

C_{2} H_{6} O_{2}

, of molar mass

2 (12.0) + 6 (1.0) + 2 (16.0) = 62 g mo l^{- 1}

, so the solution holds

83/62 = 1.339

mol in

0.625

kg of water and the molality is

m = \frac{1.339}{0.625} = 2.142 mol k g^{- 1}

Glycol is a non-electrolyte, so the van't Hoff factor is unity and

Δ T_{f} = K_{f} m = 1.86 \times 2.142 = 3.98 K

The solution freezes at

273 - 3.98 = 269.02

K, that is

269

K to the nearest integer.

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Integer answer

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