Respuesta :
Answer : The freezing point of a solution is [tex]-15.4^oC[/tex]
Explanation : Given,
Molal-freezing-point-depression constant [tex](K_f)[/tex] = [tex]5.32^oC/m[/tex]
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 :
[tex]\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}}[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
[tex]\Delta T_s[/tex] = freezing point of solution = ?
[tex]\Delta T^o[/tex] = freezing point of solvent = [tex]-10.1^oC[/tex]
i = Van't Hoff factor = 1 (for urea non-electrolyte)
[tex]K_f[/tex] = freezing point constant = [tex]5.32^oC/m[/tex]
m = molality
Now put all the given values in this formula, we get
[tex]-10.1^oC-T_s=1\times (5.32^oC/m)\times \frac{29.82g}{60.06g/mol\times 0.500kg}[/tex]
[tex]T_s=-15.4^oC[/tex]
Therefore, the freezing point of a solution is [tex]-15.4^oC[/tex]
