Question

A certain substance X has a normal freezing point of -10.1 degree C and a molal freezing point depression constant Kf = 5.32 °C.kg. mol^-1. Calculate the freezing point of a solution made of 29.82 g of urea ((NH2)2CO) dissolved in 500.

Answer 1

Answer : The freezing point of a solution is $-15.4^oC$

Explanation : Given,

Molal-freezing-point-depression constant $(K_f)$ = $5.32^oC/m$

Mass of urea (solute) = 29.82 g

Mass of solvent = 500 g  = 0.500 kg

Molar mass of urea = 60.06 g/mole

Formula used :  

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

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = ?

$\Delta T^o$ = freezing point of solvent = $-10.1^oC$

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

$K_f$ = freezing point constant = $5.32^oC/m$

m = molality

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

$-10.1^oC-T_s=1\times (5.32^oC/m)\times \frac{29.82g}{60.06g/mol\times 0.500kg}$

$T_s=-15.4^oC$

Therefore, the freezing point of a solution is $-15.4^oC$