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Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

$\begin{array}{1 1}73.43kPa\\63.43kPa\\53.43kPa\\43.43kPa\end{array} $

1 Answer

Molar mass of heptane $(C_7H_{16})=100 g mol^{-1}$
Molar mass of octane $(C_8H_{18})=114 g mol^{-1}$
$n_{neptane}=\large\frac{26g}{100g mol^{-1}}$=0.26mol
$n_{octane}=\large\frac{35g}{114g mol^{-1}}$=0.31mol
$n_{neptane}=\large\frac{0.26}{0.26+0.31}$=0.456
$n_{octane}=\large\frac{0.31}{0.26+0.31}$=0.544
$P_{neptane}=0.456\times 105.2$ kPa
$\Rightarrow 47.97$kPa
$P_{octane}=0.544\times 46.8$ kPa
$\Rightarrow 25.46$kPa
$P_{total}=47.97+25.46$
$\Rightarrow 73.43$kPa
answered Aug 5, 2014 by sreemathi.v
 
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