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

Explanation

Here, `Delta T_(f)=(273.15-271)K`
= 2.15 K
Molar mass of sugar `(C_(12)H_(22)O_(11))`
`=12xx12+22xx1+11xx16`
`= 342 g mol^(-1)`
5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 g - 5)g = 95 g of water.
Now, number of cane sugar
`= (5)/(342)=0.0146` mol
Therefore, molality of the solution,
`m=(0.0146 mol)/(0.095 kg)=0.1537 mol kg^(-1)`
Applying the relation,
`Delta T_(f)=K_(f)m`
`therefore K_(f)=(Delta T_(f))/(m)`
`= (2.15 K)/(0.1537" K kg mol"^(-1))`
= 13.99 K kg/mol
Molar of glucose `(C_(6)H_(12)O_(6))`
`= 6xx12+12xx1+6xx16`
`= 180 g mol^(-1)`
5% glucose in water means 5 g glucose is present in (100 - 5) g = 95 g of water. Therefore, Number of moles of glucose
`= (5)/(180)` mol
= 0.0278 mol
Therefore, molality of the solution,
`m =(0.0278 mol)/(0.095 kg)`
Applying the relation,
`Delta T_(f)=K_(f)xx m`
= 13.99 K kg `mol^(-1)xx 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.