Question

90 gm glucose and 120 gm urea dissolved in 1.46 kg aqueous solution, then what will be the boiling point of the solution at 1 bar pressure ? (Kb = 0.512°C -kg – mole^-1, molecular weight of glucose and urea are 180 and 60 gm/mole respectively)

(a)  100.876°C 

(b)  101.024°C

(c)  100.248°C

(d)   100.007°C 

Answer 1

Correct option (b) 101.024°C

Explanation:

mass of solvent in a solution w₀ = 1460 - (90 + 120) = 1250 gm. 

mole of glucose = 90/180 = 0.5

mole of urea = 120/60 = 2

Total moles of solute in a solution = 0.5 + 2 = 2.5