Question

. 8.00 g of a nonvolatile solute was dissolved in 200.0 g of water at 30.0 ^oC. The vapor pressure of the solution was measured and found to be 31.20 torr. The vapor pressure of pure water at 30.0 ^oC is 31.82 torr.
Calculate the molar mass of the unknown solute.

Answer 1

Let us say that the molar mass of the unknown solute = M g/mol

Mass of the unknown solute = 8.00 g

Moles of unknown solute = mass of unknown solute/molar mass of unknown solute

= 8.00/M mol

Mass of water = 200.0 g

Molar mass of water = 18 g/mol

Moles of water = mass of water/molar mass of water

= 200.0 g/18 g/mol

= 11.1 mol

Mole fraction of water in solution (X) = moles of water/(moles of unknown solute + moles of water)

= 11.1/[(8.00/M) + 11.1]

= 11.1M/(8.00 + 11.1M)

Now, from Raoult's law,

p = p^oX

where,

p = vapor pressure of the solution = 31.20 torr

p^o = vapor pressure of water (solvent) = 31.82 torr

X = mole fraction of water (solvent) = 11.1M/(8.00 + 11.1M)

Therefore,

31.20 torr = (31.82 torr x 11.1M)/(8.00 + 11.1M)

or, 249.6 + 346.32M = 353.202M

or, 6.882M = 249.6

or, M = 36.3

Hence, the molar mass of the unknown solute = 36.3 g/mol