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* m\\\\T_f^o-T_f=T_f*\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \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* \frac{5.8g* 1000}{\text{Molar mass of solute}* 100g}`

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

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