Question

60 g of glucose is dissolved in 250 g of water. Calculate the freezing point of this solution.
(molar mass of glucose = $180~g~mol^{-1}$, $K_f$ for water = $1.86~K~kg~mol^{-1}$)

Hint:

Always convert the mass of the solvent to kilograms before calculating molality.

Answer 1

Step 1: Concept
Depression in freezing point formula: $\Delta T_f = K_f \times m$.

Step 2: Meaning
$m$ is molality (moles of solute per kg of solvent).

Step 3: Analysis
Moles of glucose ($n$) = $\frac{60~g}{180~g~mol^{-1}} = \frac{1}{3}~mol$.
Mass of water ($W$) = $250~g = 0.25~kg$.
Molality ($m$) = $\frac{1/3}{0.25} = \frac{4}{3} = 1.333~m$.
$\Delta T_f = 1.86 \times \frac{4}{3} = 0.62 \times 4 = 2.48~K$.

Step 4: Conclusion
Freezing point of solution = Freezing point of pure water - $\Delta T_f$
$T_f = 273.15~K - 2.48~K = 270.67~K$ (or $-2.48^\circ C$).

Final Answer: $270.67~K$ (or $-2.48^\circ C$)