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.)

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 !

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

c8h10 is your answerr