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Depression in Freezing Point

31 g of ethyl...

Question

$31 g$ of ethylene glycol $(C_{2} H_{6} O_{2})$ is mixed with $500 g$ of solvent ( $K_{f}$ of the solvent is $2 K$ $k g$ ${m o l}^{- 1}$ ). What is the freezing point of the solution (in $K$ )?

[Freezing point of solvent $= 273 K$ ]

272

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271

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270

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274

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275

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Solution

The correct option is B $271$
Given, $W_{B} =$ mass of solute (ethylene glycol) $= 31 g$

$W_{A} =$ mass of solvent $= 500 g$

$M_{B} =$ Ethylene glycol $(C_{2} H_{6} O_{2})$

$= (2 \times 12 + 6 + 2 \times 16) = 62 g$

$K_{f} = 2 K$ $k g$ ${m o l}^{- 1}$

$Δ T_{f} = K_{f} \frac{W_{B}}{M_{B}} \times \frac{1000}{W_{A}}$

$Δ T_{f} = \frac{2 \times 31 \times 1000}{62 \times 500} = 2 K$

$∵ Δ T_{f} = T_{f}^{o} - T_{f}$

$T_{f} = T_{f}^{o} - 2$ ( $∵$ Depression in freezing point $= 2 K$ )

$∴ T_{f} = 273 - 2 = 271 K$

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Calculate the freezing point of a solution when 31 gram of Ethylene glycol is dissolved in 500 gram of water. (

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