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)

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By Expert Faculty of Sri Chaitanya
a
724
b
740
c
736
d
718
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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=724mmHg

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