Question

When the mixture of two immicible liquids (water and nitrobenzene) boils at 372 K and the vapour pressure at this temperature are $97.7kPa(H_{2}O$and $3.6kPa$ $\left(C_6{H}_{5}N{O}_2\right]).$Calculate the weight per of nitrobezene in the vapour.

• A. $10.55$
• B. $20.11$
• C. $30.84$
• D. $61.68$

Answer 1

The correct option is B $20.11$

Total pressure is equal to atmospheric pressure. It is $101.321$ kPa.
The partial pressures of water and nitrobenzene are $97.7$ kPa and $3.6$ kPa resectively.
The mole fraction of water is the ratio of the partial pressure of water to total pressure. It is
$\frac{97.7}{101.321}=0.964$
The mole fraction of nitrobenzene is the ratio of the partial pressure of nitrobenzene to total pressure. It is
$\frac{3.6}{101.321}=0.0355$
The molar masses of water and nitrobenzene are 18 g/mol and 123 g/mol respectively.
Assuming 1 mole of the mixture, the number of moles of water and nitrobenzene will be 0.964 moles and 0.0355 moles.
The mass of water will be the product of the number of moles of water to the molar mass of water. It will be
$0.964\times 18=17.36$ g
The mass of nitrobenzene will be the product of the number of moles of nitrobenzene to the molar mass of nitrobenzene. It will be
$0.0355\times 123=4.37$ g
Total mass of the mixture will be
$17.36+4.37=21.73$ g
The weight percent of nitrobenzene will be $\frac{4.37}{21.73}\times 100=20.11$ %