Question

A solution of 9.87 g of a compound having the empirical formula C6H5P in 45.0 g of benzene is observed to freeze at 2.9°C. Calculate the molar mass of the solute

Answer 1

Tf   = 5.5-2.9   = 2.6⁰C

i   = 1 for nonelectrolyte

Kf = 4.9⁰C/m

Tf = i*Kf*m

2.6 = 1*4.9*m

m     = 2.6/4.9    = 0.53m

molality    = W*1000/G.M.Wt * weight of solvent in g

0.53      = 9.87*1000/G.M.Wt * 45

G.M.Wt      = 9.87*1000/0.53*45    = 413.84g/mole

The molar mass of solute   = 413.84g/mole

The empirical formula   = C6H5P

The empirical formula of Mass = 108g/mole

molecular formula = (empirical formula)n

n          = 413.84/108   = 3.83 round off 4

molecular formula   = ( C6H5P)4

                              = C24H20P4