Answer : The molal freezing point depression constant of X is [tex]4.12^oC/m[/tex]
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 :
[tex]\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}}[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
[tex]\Delta T_s[/tex] = freezing point of solution = [tex]-0.5^oC[/tex]
[tex]\Delta T^o[/tex] = freezing point of liquid X= [tex]0.4^oC[/tex]
i = Van't Hoff factor = 1 (for non-electrolyte)
[tex]K_f[/tex] = molal freezing point depression constant of X = ?
m = molality
Now put all the given values in this formula, we get
[tex][0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}[/tex]
[tex]k_f=4.12^oC/m[/tex]
Therefore, the molal freezing point depression constant of X is [tex]4.12^oC/m[/tex]
