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

Step-by-step explanation:

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

`\Delta T_f=iK_fm`

Or,

`\Delta T_f=i* K_f* \frac{m_(solute)* 1000}{M_(solute)* 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* 5.065^oC/m* (0.500* 1000)/(M_(solute)* 50.0)\\\\M_(solute)=(1* 5.065* 0.500* 1000)/(50.0* 0.42)=120.6g/mol`

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