Answer : The molar mass of solute is, 89.9 g/mol
Explanation : Given,
Mass of solute = 5.8 g
Mass of solvent (water) = 100 g
Formula used :
[tex]\Delta T_f=K_f\times m\\\\T_f^o-T_f=T_f\times\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of water}}[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
[tex]T_f^o[/tex] = temperature of pure solvent (water) = [tex]0^oC[/tex]
[tex]T_f[/tex] = temperature of solution = [tex]1.20^oC[/tex]
[tex]K_f[/tex] = freezing point constant of water = [tex]1.86^oC/m[/tex]
m = molality
Now put all the given values in this formula, we get
[tex](0^oC)-(1.20^oC)=1.86^oC/m\times \frac{5.8g\times 1000}{\text{Molar mass of solute}\times 100g}[/tex]
[tex]\text{Molar mass of solute}=89.9g/mol[/tex]
Therefore, the molar mass of solute is, 89.9 g/mol
