Questions
For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point a 1 atm pressure is 2o C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76K kg mol-1)
detailed solution
Correct option is A
The elevation in boiling point is ΔT=Kbm;m= molality =n2w1×1000[n2 = number of moles of solute, w1= weight of solvent in gram]⇒ 2=0.76×n2100×1000⇒ n2=519Also, from Raoult's law of lowering of vapour pressure −Δpp∘=x2 −Δpp∘=n2n1+n2−Δpp∘≈n2n1 ∵n1≫n2⇒ −Δp=760×519×18100 =36mmHg⇒ p=760−36=724mmHgTalk to our academic expert!
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