Question

Show the calculation of the molar mass (molecular weight) of a solute if a solution of 5.8 grams of the solute in 100 grams of water has a freezing point of 1.20oC. Kf for water is 1.86

Answer 1

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 :  

$\Delta T_f=K_f\times m\\\\T_f^o-T_f=T_f\times\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of water}}$

where,

$\Delta T_f$ = change in freezing point

$T_f^o$ = temperature of pure solvent (water) = $0^oC$

$T_f$ = temperature of solution = $1.20^oC$

$K_f$ = freezing point constant of water = $1.86^oC/m$

m = molality

Now put all the given values in this formula, we get

$(0^oC)-(1.20^oC)=1.86^oC/m\times \frac{5.8g\times 1000}{\text{Molar mass of solute}\times 100g}$

$\text{Molar mass of solute}=89.9g/mol$

Therefore, the molar mass of solute is, 89.9 g/mol