Answer : The molality of the solution is, 0.304 mol/kg
Explanation : Given,
Molal-freezing-point-depression constant [tex](K_f)[/tex] for solvent = [tex]4.60^oC/m[/tex]
Freezing point of pure liquid = [tex]47.2^oC[/tex]
Freezing point of solution = [tex]45.8^oC[/tex]
Formula used :
[tex]\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times m[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
[tex]\Delta T_s[/tex] = freezing point of solution = [tex]45.8^oC[/tex]
[tex]\Delta T^o[/tex] = freezing point of pure liquid = [tex]47.2^oC[/tex]
i = Van't Hoff factor = 1 (for non-electrolyte)
[tex]K_f[/tex] = freezing point constant for solvent = [tex]4.60^oC/m[/tex]
m = molality
Now put all the given values in this formula, we get
[tex]47.2^oC-45.8^oC=1\times (4.60^oC/m)\times m[/tex]
[tex]m=0.304mol/kg[/tex]
Therefore, the molality of the solution is, 0.304 mol/kg
