Question

The vapor pressure of 2-propanol (C3H8O, M = 60.096 g mol–1) is 50.00 kPa at 338.8 C, but it fell to 49.62 kPa when 8.69 g of a nonvolatile organic compound was dissolved in 250 g of 2-propanol. Calculate the molar mass of the compound.

Answer 1

For vapor pressure lowering Dp = X_B p_A* , where XB is the mol fraction of solute and p_A* is the vapor pressure of pure solvent.

So           X_B = dP / P_A * = (50.00 – 49.62) kPa / 50.00 kPa = 0.0076

For vapor pressure lowering Dp = X_B p_A* , where XB is the mol fraction of solute and p_A* is the vapor pressure of pure solvent.

So           X_B =   Dp   = (50.00 – 49.62) kPa   = 0.0076

                          p_A*              50.00 kPa

But         X_B = n_B

n_A + n_B    

X_B n_A + X_B n_B = n_B

So           n_B = X_Bn_A      

                         (1 – X_B)

n_A = 250.0 g    1 mol    = 4.159 mol solvent

                         60.11 g

And so   n_B = (0.0076) (4.159 mol)   = 0.03185 mol B

                               (1. – 0.0076)

And so the molecular mass of the unknown organic compound is

                M = m_B/n_B = (8.69 g)/(0.03185 mol) = 277. g/mol

Note: If we had assumed that X_B = n_B/n_A (which is true in the limit XB = 0) in the above calculations we would have gotten M = 275. g/mol.