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The vapor pressure of benzene is 100.0 mmHg at 26.1°C. Calculate the vapor pressure of a solution containing 26.6 g of camphor (C10H16O) dissolved in 94.0 g of benzene. (Camphor is a low-volatility solid.)

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Willchemistry

Hey there!:

Molar mass benzene = 78.11 g/mol

find number of moles of benzene :

Moles of benzene =  mass / molar mass of benzene

Moles of benzene =  94.0 / 78.11

Moles of benzene = 1.203 moles

Molar mass camphor =  152.23 g/mol

Moles of camphor = 26.6 / 152.23

Moles of camphor = 0.1747 moles

Therefore , vapor pressure of solution is :

mole fraction of benzene * vapor pressure of solution

= 1.203 / ( 1.203 + 0.1747 ) * 100.0

=  ( 1.230 / 1.3777 ) * 100.0

= 0.8927 * 100.0

=> 89.27 mmHg

Hope this helps !

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