Answer: [tex]-15.4^00C[/tex]
Explanation:
[tex]T^0_f-T_f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]
where,
[tex]T_f[/tex] = boiling point of solution = ?
[tex]T^o_f[/tex] = boiling point of solvent (X) = [tex]-10.1^oC[/tex]
[tex]k_f[/tex] = freezing point constant = [tex]5.32^oC/kgmol[/tex]
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
[tex]M_2[/tex] = molar mass of solute (urea) = 60 g/mol
Now put all the given values in the above formula, we get:
[tex](-10.1-(T_f))^oC=1\times (5.32^oC/m)\times \frac{(29.82g)\times 1000}{60\times (500.0g)}[/tex]
[tex]T_f=-15.4^0C[/tex]
Therefore, the freezing point of solution is [tex]-15.4^0C[/tex]
