Question

A 1.8 g mass of fructose is added to 0.100 kg of water and it is

Answer 1

Answer : The molar  mass of fructose is 180

Explanation : Given,

Molal-freezing-point-depression constant $(K_f) = 1.86^oC/m$

Mass of fructose (solute) = 1.8 g

Mass of water (solvent) = 0.100 kg

Formula used :

$\Delta T_f=i\times K_f\times m\\\\\Delta T_f=i\times K_f\times\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}\times \text{Mass of solvent in Kg}}$

where,

$\Delta T_f$ = change in freezing point  = $0.186^oC$

i = Van't Hoff factor = 1

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

m = molality

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

$0.186^oC=1\times (1.86^oC/m)\times \frac{1.8g}{\text{Molar mass of fructose}\times 0.100kg}\\\\\text{Molar mass of fructose}=180g/mol$

Therefor, the molar  mass of fructose is 180