Question

A. 0.2436 sample of an unknown substance was dissolved in 20.0mL of cyclohexane. The density of cyclohexane is 0.779 g/mL. The freezing point depression was 2.50 oC and the Kf value for cyclohecane is 20.5oC/m. Calculate the molality of the above solution and the molar mass of unknown.

Answer 1

Answer: The molality of solution is 0.122 m and molar mass of unknown substance is 128.16 g/mol

Explanation:

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

$\Delta T_f=iK_fm$

where,

$\Delta T_f$ = depression in freezing point = 2.50°C

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

$K_f$ = molal freezing point elevation constant = 20.5°C/m

m = molality of solution = ?

Putting values in above equation, we get:

$2.50^oC=1\times 20.5^oC/m\times m\\\\m=\frac{2.50}{1\times 20.5}=0.122m$

• To calculate the mass of cyclohexane, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of cyclohexane = 0.779 g/mL

Volume of cyclohexane = 20.0 mL

Putting values in above equation, we get:

$0.779g/mL=\frac{\text{Mass of cyclohexane}}{20.0mL}\\\\\text{Mass of cyclohexane}=(0.779g/mL\times 20.0mL)=15.58g$

• To calculate the molar mass of substance for given molality, we use the equation:

$\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$m_{solute}$ = Given mass of solute = 0.2436 g

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

$W_{solvent}$ = Mass of solvent (cyclohexane) = 15.58 g

Molality of solution = 0.122 m

Putting values in above equation, we get:

$0.122=\frac{0.2436\times 1000}{M_{solute}\times 15.58}\\\\M_{solute}=\frac{0.2436\times 1000}{0.122\times 15.58}=128.16g/mol$

Hence, the molality of solution is 0.122 m and molar mass of unknown substance is 128.16 g/mol