$5.8 g$ of non volatile solute was dissolved in 100 g of carbon disulphide $(C S_{2})$ . The vapour pressure of the solution was found to be 190 mm. of Hg. Calculate the molar mass of the solute given the vapour pressure of pure $C S_{2}$ is 195 mm. of Hg [Molar mass of $C S_{2} = 76 g m o l^{- 1}$ ].

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Solution

The relative lowering in the vapour pressure is equal to the mole fraction of solute. $P^{0}$ and P are the vapour pressures of pure solute and solution respectively. $W_{1}$ and $M_{1}$ are mass and molar mass of solvent respectively. $W_{2}$ and $M_{2}$ are mass and molar mass of solute respectively.
$\frac{P^{0} - P}{P^{0}} = \frac{\frac{W_{2}}{M_{2}}}{\frac{W_{2}}{M_{2}} + \frac{W_{1}}{M_{1}}}$
$\frac{195 - 190}{195} = \frac{\frac{5.8}{M_{2}}}{\frac{5.8}{M_{2}} + \frac{100}{76}}$
$0.0256 = \frac{\frac{5.8}{M_{2}}}{\frac{5.8}{M_{2}} + 1.32}$
Taking reciprocal on both sides
$\frac{1}{0.0256} = \frac{\frac{5.8}{M_{2}} + 1.32}{\frac{5.8}{M_{2}}}$
$39 = \frac{\frac{5.8}{M_{2}} + 1.32}{\frac{5.8}{M_{2}}}$
$39 = 1 + \frac{1.32}{\frac{5.8}{M_{2}}}$
$38 = \frac{1.32}{\frac{5.8}{M_{2}}}$
$\frac{5.8}{M_{2}} = \frac{1.32}{38} = 0.03474$
$M_{2} = 167$ g/mol

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