Question

The molecular weight of an organic compound was determined by measuring the freezing point depression of a benzene solution. A 0.500 g sample was dissolved in 50.0 g of benzene, and the resulting depression was 0.42°C (kf (benzene) = 5.065°C/m). What is the approximate molecular weight of the compound?

Answer 1

Answer: The molar mass of the sample is 120.6 g/mol

Explanation:

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$\Delta T_f$ = depression in freezing point = 0.42°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 5.065°C/m

$m_{solute}$ = Given mass of solute (sample) = 0.500 g

$M_{solute}$ = Molar mass of solute (sample) = ? g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 50.0 g

Putting values in above equation, we get:

$0.42^oC=1\times 5.065^oC/m\times \frac{0.500\times 1000}{M_{solute}\times 50.0}\\\\M_{solute}=\frac{1\times 5.065\times 0.500\times 1000}{50.0\times 0.42}=120.6g/mol$

Hence, the molar mass of the sample is 120.6 g/mol