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

Step-by-step 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* 20.5^oC/m* m\\\\m=(2.50)/(1* 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* 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)* 1000}{M_(solute)* 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=(0.2436* 1000)/(M_(solute)* 15.58)\\\\M_(solute)=(0.2436* 1000)/(0.122* 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