Question

The normal freezing point of a certain liquid

Answer 1

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

Explanation :  Given,

Mass of urea (solute) = 5.90 g

Mass of X liquid (solvent) = 450.0 g

Molar mass of urea = 60 g/mole

Formula used :  

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}\times 1000}{\text{Molar mass of urea}\times \text{Mass of X liquid}}$

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 of X = ?

m = molality

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

$[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}$

$k_f=4.12^oC/m$

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