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A solution of sucrose in 100 g of water boils at 100.25 °C. If this solution is boiled till its temperature is 101 °C, the mass of water evaporated from the solution is

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50g

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detailed solution

Correct option is C

We have from $Δ T_{b} = m K_{b}$

$\frac{{(Δ T_{b})}_{2}}{{(Δ T_{b})}_{1}} = \frac{m_{2}}{m_{1}} = \frac{n_{2} / {(m_{water})}_{2}}{n_{2} {(m_{water})}_{1}} = \frac{{(m_{water})}_{1}}{{(m_{water})}_{2}}$

Hence, ${(m_{water})}_{2} = {(m_{water})}_{1} {\{Δ T_{b})}_{1} / {(Δ T_{b})}_{2}\} = (100 g) (0.25 K / 1 K) = 25 g .$

Mass of water evaporated $= 100 g - 25 g = 75 g$

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