What is the melting point of $0.2 L$ water containing $6.20 g$ $C_{2} H_{6} O_{2}$ ?

- 1.86

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- 0.93

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0

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0.93

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1.86

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Solution

The correct option is D $- 0.93$
The molar mass of $C_{2} H_{6} O_{2} = 2 (12) + 6 (1) + 2 (16) = 62 g / m o l$
The number of moles of
$C_{2} H_{6} O_{2} = \frac{6.20 g}{62 g / m o l} = 0.1 m o l$
The molarity of $C_{2} H_{6} O_{2} = \frac{0.1 m o l}{0.2 L} = 0.5 M$
For dilute solutions, the molality is equal to molarity.
The depression in freezing point
$Δ T_{f} = K_{f} \times m$
$Δ T_{f} = 1.86 \times 0.5$
$Δ T_{f} = 0.93$ .
The melting point of pure water is 0 deg C.
The melting point of solution will be $0 - 0.93 = - 0.93^{o} C$ .

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