Question

A \(5\%\) solution (by mass) of cane sugar in water has freezing point of 271K.Calculate the freezing point of \(5\%\) glucose in water if freezing point of pure water is 273.15 K.

Answer 1

The correct answer is: 269.06 K
Here, \(ΔT_f = (273.15 - 271) K \)
\(= 2.15 K\)
Molar mass of sugar \((C_{12}H_{22}O_{11}) = 12 × 12 + 22 × 1 + 11 × 16 \)
\(= 342 g mol ^{- 1}\)
\(5\%\) solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 - 5)g = 95 g of water.
Now, number of moles of cane sugar = \(\frac{5}{342} mol\)
= 0.0146 mol 
Therefore, molality of the solution, \(m = \frac{0.0146mol}{0.095kg}\)
\(= 0.1537 mol kg^{-1}\)
Applying the relation,
\(ΔT_f = K_f × m\)
\(⇒K_f =\frac{ΔT_f }{m}\)
\(=\frac{2.15 K}{(0.1537)molkg^{-1}}\)
\(= 13.99 K\, kg mol^{- 1 }\)
Molar of glucose \((C_6H_{12}O_6) = 6 × 12 + 12 × 1 + 6 × 16\)
\(= 180 g mol^{-1}\)
\(5\%\) glucose in water means 5 g of glucose is present in (100 - 5) g = 95 g of water.
∴ Number of moles of glucose \(=\frac{5}{180} mol\)
=0.0278 mol 
Therefore, molality of the solution, \(m= \frac{0.0278mol}{0.095kg}\)
\(= 0.2926\, mol kg^{-1 }\)
Applying the relation, 
\(ΔT_f = K_f × m \)
\(= 13.99 K\, kg mol^{-1} \times 0.2926\, mol kg^{-1 }\)
= 4.09 K (approximately) 
Hence, the freezing point of 5% glucose solution is (273.15 - 4.09) K= 269.06 K.