SolutionHard
Question
A solution containing 30 g of a non-volatile solute exactly in 90 g water has a vapour pressure of 2.8 kPa at 298 K. On adding 18 g of water in the solution, the new vapour pressure becomes 2.9 kPa at 298 K. The molecular mass of the solute is
Options
A.23
B.69
C.34
D.43
Solution
$P = X_{2}.P^{o}$
$2.8 = \frac{90/18}{\frac{30}{M} + \frac{90}{18}} \times P^{o}\text{ and }2.9 = \frac{108/18}{\frac{30}{M} + \frac{108}{18}}.P^{o} $$$\therefore M = 23$$
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