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.