The mass of non-volatile, non-electrolyte solute (molar mass $50 g {m o l}^{- 1}$ ) needed to be dissolved in $114 g$ octane to reduce its vapour pressure to $75$ %, is ___________.

150 g

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75 g

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37.5 g

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50 g

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Solution

The correct option is A $150 g$
Octane has molar mass of 114 g/mol.

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

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

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

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

$W_{2} = 150$ g

Note : $W_{2}$ and $M_{2}$ are mass and molar mass of solute and $W_{1}$ and $M_{1}$ are mass and molar mass of octane.

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The mass of non-volatile, non-electrolyte solute (molar mass 50gmol−1) needed to be dissolved in 114g octane to reduce its vapour pressure to 75%, is ___________.

The correct option is A 150gOctane has molar mass of 114 g/mol.ΔPP=W2M2W2M2+W1M175100=W250g/molW250g/mol+114g114g/mol0.75=W250W250+1W250+1=W250×0.75W2=150 gNote : W2 and M2 are mass and molar mass of solute and W1 and M1 are mass and molar mass of octane.

The mass of non-volatile, non-electrolyte solute (molar mass $50 g {m o l}^{- 1}$ ) needed to be dissolved in $114 g$ octane to reduce its vapour pressure to $75$ %, is ___________.