Question

A compound weighing 0.458 g is dissolved in 30.0 g of acetic acid. The freezing point of the solution is found to be 1.50 K below that of the pure solvent. Calculate the molar mass of the compound.

Answer 1

Answer:

Molar mass of compound = 38.17 g/mol

Explanation:

The mass of compound dissolved = 0.458 g

The mass of acetic acid taken = 30.0g = 0.03 kg

the depression in freezing point =1.50 K or 1.50 ⁰C

the relation between depression in freezing point and molality is:

Depression in freezing point = Kf X molality

Where Kf= cryoscopic constant = 3.90 ⁰C Kg/mol

Putting values

1.50 = 3.90 X molality

$molality=\frac{1.50}{3.90}=0.385$

molality is moles of solute per Kg of solvent

$molality=\frac{moles}{massofsolvent}=\frac{moles}{0.03}=0.385$

moles = 0.385 X 0.03 = 0.012

$moles=\frac{mass}{molarmass}$

$molarmass=\frac{mass}{moles}=\frac{0.458}{0.012}= 38.17g/mol$

Answer 2

Answer: The molar mass of the compound is 39.69 g/mol

Explanation:

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

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 = 1.50 K = 1.50°C   (Change remains constant)

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

$K_f$ = molal freezing point elevation constant = 3.90°C/m

$m_{solute}$ = Given mass of solute = 0.458 g

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

$W_{solvent}$ = Mass of solvent (acetic acid) = 30.0 g

Putting values in above equation, we get:

$1.50^oC=1\times 3.90^oC/m\times \frac{0.458\times 1000}{\text{Molar mass of solute}\times 30.0}\\\\\text{Molar mass of solute}=\frac{1\times 3.90\times 0.458\times 1000}{1.50\times 30.0}=39.69g/mol$

Hence, the molar mass of the compound is 39.69 g/mol