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If 6.2 g of ethylene glycol (C₂H₆O₂) is mixed with 500 g of water. The freezing point of the solution obtained will be (K_f for water = 1 .86 K kg mol^-1)

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169.5 K

272.63 K

270.95 K

373.372 K

answer is B.

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Detailed Solution

$\begin{array}{l} {ΔT}_{f} = K_{f} m \\ 273 - T_{f} = 1 .86 \times \frac{6 .2 \times 1000}{62 \times 500} \\ T_{f} = 273 - 0 .372 = 272 .63 K \end{array}$

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