Question

Physical Chemistry

1.At 50°C the vapor pressure of ethyl alcohol is 219.9 mm Hg. If 6 g of a nonvolatile solute of molecular
weight 120 are dissolved in 150 g of water and ethyl alcohol, what will be the relative vapor pressure
lowerings in the two solvents?

2.A solution composed of 10 g of a nonvolatile organic solute in 100 g of diethyl ether has a vapor pressure
of 426.0 mm at 20 °C. If the vapor pressure of the pure ether is 442.2 mm at the same temperature,
what is the molecular weight of the solute?

Answer 1

1. Molecular weight of ethyl alcohol is 46g.

Therefore no.of moles in 150g ethyl alcohol is 150÷46=3.26 moles

No.of moles of non volatile solute added is wt ÷mol.wt = 6÷ 120

=0.05moles

Mole fraction of in the mixture is .05 ÷ (3.26+.05)

That is mole fraction is 0.0151

Relative lowering of vapour pressure is mol fraction of solute × vapour pressure of ethyl alcohol

That is .0151× 219.9= 3.32 mm Hg

For water solution

Moles of water in 150 g is 150÷18=8.33

Moles of solute is .05

Mol fraction of solute is (.05÷(8.33+.05))=.00596

Lowering of vapour pressure is 0.00596×92.5=.5517mm Hg

Vapour pressure of water at 50deg celsius is 92.5 mm Hg

2.Mol fraction ×vapour pressure =vapour pressure of mixture

Moles of diethyl ether is 100÷ 76= 1.31

Let mol fraction be 'a'

a × 442.2 =426

Therefore a is equal to .9638

Let moles of organic solute be' y'

Then mol fraction a = 1.31÷(y +1.31)

Solving we get y is equal to .049moles

10g of solute is .049 moles , so molecular weight is 10÷.049=204.08 g