Question

An aqueous solution containing 16.5 g of an unknown molecular (nonelectrolyte) compound in 106.0 g of water was found to have a freezing point of -1.9 ∘C. Calculate the molar mass of the unknown compound.

Answer 1

Answer : The molar mass of unknown compound is 152.38 g/mole

Explanation :

Depression in freezing point = $1.9^oC$

Mass of unknown compound = 16.5 g

Mass of water = 106.0 g

Formula used :  

$\Delta T_f=i\times K_f\times m\\\\\Delta T_f=i\times K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of water in Kg}}$

where,

$\Delta T_f$ = depression in freezing point

i = Van't Hoff factor = 1 (for non-electrolyte)

$K_f$ = freezing point constant for water = $1.86^oC/m$

m = molality

Now put all the given values in this formula, we get

$1.9^oC=1\times (1.86^oC/m)\times \frac{16.5g\times 1000}{\text{Molar mass of unknown compound}\times 106.0g}$

$\text{Molar mass of unknown compound}=152.38g/mole$

Therefore, the molar mass of unknown compound is 152.38 g/mole

Answer 2

Answer: The molar mass of the unknown compound is 152 g/mol.

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f-^0T_f=(0-(-1.9)^0C=1.9^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant = $1.86^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (water)= 106.0 g = 0.106 kg

Molar mass of unknown non electrolyte = M g/mol

Mass of unknown non electrolyte added = 16.5 g

$1.9=1\times 1.86\times \frac{16.5g}{M g/mol\times 0.106kg}$

$M=152g/mol$

Thus the molar mass of the unknown compound is 152 g/mol.