The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol - 1 ) needed to be dissolved in 114 g octane to reduce its vapour pressure by 75 %, is:

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TestHub · Accepted Answer

Molar mass of octane = 114 g/mol.

From the lowering of vapour pressure, we have,

$\frac{∆ P}{P} = \frac{\frac{W_{2}}{M_{2}}}{\frac{W_{2}}{M_{2}} + \frac{W_{1}}{M_{1}}}$

Where $W_{2} a n d M_{2}$ are mass and molar mass of solute
and $W_{1} a n d M_{1}$ are mass and molar mass of octane.

$\frac{75}{100} = \frac{\frac{W_{2}}{50 g / m o l}}{\frac{W_{2}}{50 g / m o l} + \frac{114 g}{114 g / m o l}}$

$0.75 = \frac{\frac{W_{2}}{50}}{\frac{W_{2}}{50} + 1}$

$\frac{W_{2}}{50} + 1 = \frac{W_{2}}{50 \times 0.75}$

$W_{2} = 150 g$

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