Question

10. A 5% aqueous solution (by mass) of cane
sugar (molar mass 342 g/mol) has freezing
point of 271K. Calculate the freezing point
of 5% aqueous glucose solution. (269.06
K)​

Answer 1

Answer:

if its wrong you can comment it openly or it might help you out  : ) happy to help.

In case of cane sugar:

ΔTf = (273.15 - 271) K = 2.15 K

Molar mass of sugar (C12H22O11) = 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 =5/342 mol

= 0.0146 mol

Therefore, molality of the solution,m =0.0146mol / 0.095kg  

= 0.1537 kg mol - 1

Now applying the relation,

ΔTf = Kf × m

⇒ Kf =  ΔTf  / m

⇒  2.15K / 0.1537 kg mol-1

= 13.99 K kg mol-1

Molar of glucose (C6H12O6) = 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.

Therefore  number of moles of glucose = 5/180 mol

= 0.0278 mol .

Therefore, molality of the solution,m =0.0278 mol / 0.095 kg

= 0.2926 mol kg - 1

Applying the relation,

ΔTf = Kf × m

= 13.99 K kg mol - 1 × 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.