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Reaction of Metals with Acid

Dry air was p...

Question

Dry air was passed through bulbs containing a solution of $40 g$ of non-volatile electrolytic solute in $360 g$ of water, then through bulbs containing pure water at the same temperature and finally through a tube in which pumice moistened with strong $H_{2} S O_{4}$ was kept. The water bulbs lost $0.0870 g$ and the sulphuric acid tube gained $2.175 g$ . Calculate the molar mass of solute (in gram)

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Solution

By using the formula,
$\frac{P - P_{S}}{P} = x_{s o l u t e}$
$x=Mole fraction of solute=Number of moles of soluteTotal number of moles$

Substituting the values, we get
$\frac{0.087}{2.175} = \frac{\frac{40}{M}}{\frac{40}{M} + \frac{360}{18}}$
$\Rightarrow 25 = 1 + \frac{M}{2}$
$\Rightarrow M = 48 g$

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