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Lelu
2 months ago
14

The vapor pressure of benzene, c6h6, is 100.0 torr at 26.1 °c. Assuming raoult's law is obeyed, how many moles of a nonvolatile

solute must be added to 100.0 ml of benzene to decrease its vapor pressure by 10.0% at 26.1 °c? The density of benzene is 0.8765 g>cm3.
Chemistry
1 answer:
Anarel [2.9K]2 months ago
3 0

By Raoult's law:

p_{solution} = X_{solvent}\times P_{solvent}    -(1)

Where p_{solution} represents the vapor pressure measured for the solution, X_{solvent} denotes the mole fraction of the solvent, and P_{solvent} specifies the vapor pressure of the pure solvent.

The vapor pressure for benzene is recorded as 100 torr at 26.1^{o} C   (as given).

Adding a non-volatile solute decreases the vapor pressure by 10.0%   (as given).

The vapor pressure of the solution component is thus given by 100 - 10 = 90 torr.

Substituting the known values into formula (1):

90 = mole fraction of benzene \times 100

mole fraction of benzene = \frac{90}{100} = 0.9

The number of moles of benzene is Volume\times \frac{density}{Molar Mass}.

Calculating to find moles of benzene yields 100 cm^{3}\times \frac{0.8765 g/cm^{3}}{78 g/mol} = 1.124 mol.

Since, mole fraction of benzene =\frac{mole of benzene }{mole of benzene +mole of non volatile solute }

Thus, 0.9 =\frac{1.124 }{1.124+mole of non volatile solute }

\frac{1.124}{0.9} =1.124+mole of non volatile solute

mole of non volatile solute = 1.249 - 1.124 = 0.125

Therefore, the moles of non-volatile solute required is 0.125.

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