An aqueous sol: contaning 10g of mixture of urea and glucose boils at 100.58°C. Addition of a further 6.0g glucose to the above solution causes it to boil at 100.77° C mass percentage of urea in the original mixture is 7y then y = ?

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answer is 2.

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Detailed Solution

$\begin{array}{l} {ΔT}_{b} = i . m \times K_{b} \\ 0 .58 = 1 \times [\frac{x}{60} + \frac{y}{180}] \times \frac{100}{W_{A}} \times K_{b} \\ 0 .77 = 1 \times [\frac{x}{60} + (\frac{y + 6}{180})] \times \frac{100}{W_{A}} \times K_{b} \\ \Rightarrow [\frac{0 .77}{0 .58}] = \frac{\frac{x}{60} + (\frac{y + 6}{180})}{(\frac{x}{60} + \frac{y}{180})} \\ = 1 .328 (\frac{10 - y}{60}) + \frac{1 .328 y}{180} \\ (\frac{10 - y}{60}) + (\frac{y + 6}{180}) - 1 .328 (\frac{10 - y}{60}) - (\frac{1 .328 y}{180}) = 0 \begin{array}{l} \frac{3 y}{180} - \frac{0 .328 y}{180} + \frac{1}{30} - \frac{0 .328 \times 5}{30} = 0 \\ y (\frac{3 - 0 .328}{180}) + (\frac{1 - 1 .64}{30}) = 0 \end{array} \end{array}$

y =2

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