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)

Question

Infinity Learn · Accepted Answer

The elevation in boiling point is           Δ$$T=Kbm$$;$$m=$$ 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∘≈$$n2n1$$                         ∵$$n1$$≫$$n2$$⇒   −Δ$$p=760$$×$$519$$×$$18100$$                 $$=36mmHg$$⇒      $$p=760$$−$$36=724mmHg$$

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