Question

if 0.78g benzene is dissolved 0.9g of toluene so find out the partial pressure of toluene and benzene and total vapour pressure p0B=400 mm hg
p0T=200 mm hg

Answer 1

Given:0.78g benzene is dissolved 0.9g of toluene

p0B=400 mm hg

p0T=200 mm hg

To Find: partial pressure of toluene and benzene and total vapor pressure

Explanation:

$n= \frac{given mass}{molar mass}$

where n is the number of moles

moles of benzene: $n_{b}=\frac{0.78}{78} = 0.01$

moles of toluene: $n_{t}= \frac{0.9}{92} =0.009$

mole fraction of benzene: $x_{b} = \frac{n_{b} }{n_{b}+n_{t} } =\frac{0.01}{0.01+0.009} =0.52$

mole fraction of toluene:$x_{t} = \frac{n_{t} }{n_{b}+n_{t} } =\frac{0.009}{0.01+0.009} =0.47$

Partial pressure(pb) is the product of vapor pressure of pure substance and its mole fraction.

$p_{b} = P_{b}\times x_{b}\\p_{b} = 400\times0.52 = 208$

Hence, partial pressure of benzene is 208 mmHg.

In similar way, partial pressure of toluene can be calculated:

$p_{t} = P_{t}\times x_{t}\\p_{t} = 200\times0.47 = 94$

partial pressure of toluene is 94 mmHg.

Total vapor pressure of the system:

$P_{total} = p_{b} + p_{t}\\=208+94\\=302 mm Hg$

Total vapor pressure of the system is 302 mmHg.