Question

At 25^°C, the vapor pressure of benzene is 96.0 mm Hg; 30.3 mm Hg for toluene, and 23.76 mm hg).

1. What is:

(i) the vapor pressure of a mixture of 10.0 g of benzene and 20.0 g toluene at 25 ^oC, and;

(ii) the mole fraction benzene and toluene in the vapor of the solution?

2. What is the vapor pressure of a solution of 15.0 g of NaNO₃ in 50.0 g of water at 25^°

^{Please answer all questions}

^{Thank You}

Answer 1

As per Raoults law

Pa = Xa x Pao

mixture of 10.0 g of benzene and 20.0 g toluene

10/78g/mol = 0.128 mole

20 /92g/mol = 0.217 mole

So mole fraction of benzene = 0.128/(0.128 + 0.217) = 0.37

mole fraction of toluene = 0.217/(0.128 + 0.217) = 0.628

Total vapour pressure = 96 x 0.37 + 30.3 x 0.628

= 54.55 mm Hg

b) mole fraction of benzene in vapour is 96 x 0.37/54.55 = 0.651

mole fraction of toluene in vapour = 30.3 x 0.628/54.55 = 0.349

c) 15 g of NaNO3 = 15/85g/mol = 0.176 moles

50 g of water is 50/18g/mol = 2.778 moles

mole fraction of water = 2.778/(2.778 + 0.176) = 0.94

Vapour pressure of the solution = 0.94 x 23.76 = 22.34 mm Hg