Question

A certain substance X has a normal freezing point of -10.1 degree C and a molal freezing point depression constant K_f = 5.32 degree C middot kg middot mol^-1. Calculate the freezing point of a solution made of 29.82 g of urea ((NH_2)_CO) dissolved in 500. g of X. Be sure your answer has the correct number of significant digits.

Answer 1

Answer: $-15.4^00C$

Explanation:

$T^0_f-T_f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = boiling point of solution = ?

$T^o_f$ = boiling point of solvent (X) = $-10.1^oC$

$k_f$ = freezing point constant  = $5.32^oC/kgmol$

m = molality

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

= mass of solute (urea) = 29.82 g

= mass of solvent (X) = 500.0 g

$M_2$ = molar mass of solute (urea) = 60 g/mol

Now put all the given values in the above formula, we get:

$(-10.1-(T_f))^oC=1\times (5.32^oC/m)\times \frac{(29.82g)\times 1000}{60\times (500.0g)}$

$T_f=-15.4^0C$

Therefore, the freezing point of  solution is $-15.4^0C$