Question

The normal freezing point of a certain liquid

X

is

0.4°C

, but when

5.90g

of urea

NH22CO

are dissolved in

450.g

of

X

, it is found that the solution freezes at

−0.5°C

instead. Use this information to calculate the molal freezing point depression constant

Kf

of

X

.
Be sure your answer has the correct number of significant digits.

Answer 1

Answer : The molal freezing point depression constant of liquid X is, $4.12^oC/m$

Explanation :  Given,

Mass of urea (solute) = 5.90 g

Mass of liquid X (solvent) = 450 g  = 0.450 kg

Molar mass of urea = 60 g/mole

Formula used :  

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}}{\text{Molar mass of urea}\times \text{Mass of liquid X Kg}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = $-0.5^oC$

$\Delta T^o$ = freezing point of liquid X = $0.4^oC$

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

$K_f$ = Molal-freezing-point-depression constant = ?

m = molality

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

$0.4^oC-(-0.5^oC)=1\times K_f\times \frac{5.90g}{60g/mol\times 0.450kg}$

$K_f=4.12^oC/m$

Therefore, the molal freezing point depression constant of liquid X is, $4.12^oC/m$