SolutionHard
Question
Air was drawn through a solution containing 40 g of solute (non-volatile and non-electrolyte) in 100 g of water and then through water. The loss of mass of water was 0.05 g and the total mass of water absorbed in sulphuric acid tube was 2.05 g. The molecular mass of the solute is
Options
A.267.86
B.288.0
C.295.2
D.302.4
Solution
Loss is mass of solvent, w1 $\propto$ (P° – P) and gain is mass of absorbent, w2 $\propto$ P°
$\therefore\frac{w_{1}}{w_{2}} = \frac{P^{o} - P}{P^{o}} = X_{1} \Rightarrow \frac{0.05}{2.05} = \frac{40/M}{\frac{40}{M} + \frac{100}{18}} \Rightarrow M = 288$
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