Question

A 1.8 g mass of fructose is added to 0.100 kg of water and it is
found that the freezing point has decreased by 0.186 °C. Given
that the Kf value of water is 1.86 °C kg/mol, what is the molar
mass of fructose (van't Hoff factor, i = 1)?
g/mol
Round your answer to the nearest whole number. Do not include units in your answer​

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

Answer 2

The molar mass of fructose will be "180 g/mol". To understand the calculation check below.

Molar mass and Molality

According to the question,

Fructose mass = 1.8 g

Water's mass = 0.100 kg

Molal freezing point depression constant, $K_f$ = 1.86°C/m

Freezing point change, Δ$T_f$ = 0.186°C

Freezing point constant, $K_f$ = 1.86°C/m

We know the relation,

→ Δ$T_f$ = i × $K_f$ × m

or,

         = i × $K_f$ × $\frac{Fructose \ mass}{Fructose \ molar \ mass\times Solvent's \ mass}$

By substituting the values, we get

0.186 = 1 × 1.86 × $\frac{1.8}{Fructose \ molar \ mass\times 0.100}$

By applying cross-multiplication,

Molar mass = 180 g/mol

Thus the above approach is right.          

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