Question

1539molal aqueous solutionof cane sugar has freezing point of 271 k while the freezing point of pure water is 273.15k what will be freezing point of an aqueous solutioncontaing5g glucose per 100g of solution

Answer 1

Answer:

Explanation:

Freezing point of water = 217k

Glucose in water = 5%

Molar mass of cane sugar C12H22O11 =342gmol−1

Molarity of sugar = 5×1000/342×100

= 0.146

ΔTf for sugar solution = 273.15 - 271  

= 2.15°

Since, ΔTf = Kf × m

Kf = 2.15/0.146

Molarity of glucose solution = 5/180 × 1000/100  

= 0.278

ΔTf(Glucose) = 2.150/146 × 0.278

= 4.09°

Freezing point = 273.15−4.09°

= 269.06K

Therefore, the freezing point is 269.06K