Calculate the mass of a non-volatile solute (molar mass $40 g m o l^{- 1}$ ) which should be dissolved in $114 g$ octane to reduce its vapour pressure to $80$ %.

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Solution

Let $p$ be the vapour pressure of pure octane. The vapour pressure of solution will be $\frac{80}{100} p = 0.8 p$ .
Molar mass of solute (M) and octane (m) are $40$ g/mol and $114$ g/mol respectively. Mass of octane, w is $114$ g.
$\frac{p - p^{'}}{p} = \frac{W m}{M w} \frac{p - 0.8 p}{p} = \frac{W \times 114}{40 \times 114} W = 8 g$
Hence, $8$ g of solute are required.

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Calculate the mass of a non-volatile solute (molar mass 40 g mol−1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Let p be the vapour pressure of pure octane. The vapour pressure of solution will be 80100p=0.8 p.Molar mass of solute (M) and octane (m) are 40 g/mol and 114 g/mol respectively. Mass of octane, w is 114 g.p−p′p=WmMwp−0.8pp=W×11440×114W=8gHence, 8 g of solute are required.

Calculate the mass of a non-volatile solute (molar mass $40 g m o l^{- 1}$ ) which should be dissolved in $114 g$ octane to reduce its vapour pressure to $80$ %.