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Relative Lowering of Vapour Pressure

C6H6 freezes ...

Question

$C_{6} H_{6}$ freezes at ${5.5}^{o} C$ . At what temperature will a solution of 10.44g of $C_{4} H_{10}$ in 200 g of $C_{6} H_{6}$ freeze?
$K_{f} (C_{6} H_{6}) = {5.12}^{o} C / m$

{4.608}^{o} C

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{0.892}^{o} C

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{5.5}^{o} C

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None of the above

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Solution

The correct option is A ${4.608}^{o} C$
Given $= K_{f} (C_{6} H_{6}) = 512^{0} c / m$ $|$ Molar mass of $C_{4} H_{10} = 58$
$10.44$ g of $C_{4} H_{10}$ is dissolved in $200$ g of $C_{6} H_{6}$ .
$∴$ Molality $= \frac{10.44 / 58}{200 / 1000} = 0.9 M$
$∴$ ${Δ T}_{f} = K_{f} . M$
$= 5.12 \times 0.9$
$= {4.608}^{0} C$

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