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A radiator of motor vehicle was filled with 8.2L of water to which 2L of methanol (d=0.8gm/ml) were added, the lowest temp. at which the vehicle can be parked without a danger of getting water in the radiator to freeze is (K_f for water = 1.86 Km^–1)

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–5.6°C

–6.6°C

–4.3°C

–9.5°C

answer is A.

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Detailed Solution

$\begin{array}{l} {ΔT}_{f} = K_{f} \times \frac{Wt}{Gmt} \times \frac{1000}{Wt . of Sovlent} \\ = 1 .86 \times \frac{1600}{32} \times \frac{1000}{8200} \end{array}$

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