Question

One mole of a nonelectrolyte was dissolved in 250 g of water and the solution heated to 100C. Given that the vapor pressure of water at 100C is 760 torr, what is the vapor pressure of the solution at this temperature. Show your calculations please!

Answer 1

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 250 g

number of mol of H2O,

n = mass/molar mass

=(250.0 g)/(18.016 g/mol)

= 13.88 mol

we have:

n(water),n1 = 13.88 mol

n(non electrolyte),n2 = 1 mol

Total number of mol = n1+n2

= 13.88 + 1

= 14.88 mol

Mole fraction of each components are

X(water) = n1/total mol

= 13.88/14.88

= 0.9328

According to Raoult’s law:

P = Po*X(solvent)

p = 760.0*0.9328

p = 709 torr

Answer: 709 torr