3. For a dilute solution containing...
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3. 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 at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take K6 = 0.76 K kg mol-'). (2012) (a) 724 (b) 740 (c) 736 (d) 718

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Chemistry
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( Delta mathrm{Tb}=mathrm{K}_{mathrm{b}} mathrm{m} ) ( mathrm{m}= ) molality of solute in the solvent ( mathrm{K}_{mathrm{b}}= ) ebuloscopic constant ( (text { it is not given in question, So I assume it 0.76 } mathrm{K} mathrm{Kg} / mathrm{mol} ) ( mathrm{m}=frac{Delta mathrm{Tb}}{mathrm{Kb}}=frac{2 mathrm{K}}{0.76} ) ( mathrm{m}=frac{mathrm{n}_{2}(text { amount of solute })}{mathrm{m}_{1}(text { mass of solvent in } mathrm{Kg})} ) ( mathrm{n} 2=mathrm{mm}_{1}=left(frac{2}{0.76}right)(0.1)=frac{2}{7.6} ) From expression, relative lowering in vapour pressure, ( Delta p=x 2 p_{1}^{0} ) ( n mid>>>n 2 ) ( x_{2}=frac{n_{2}}{n_{1}+n 2}=frac{n_{2}}{n_{1}}=frac{n_{2}}{m_{1} / M_{1}}=frac{2 / 0.76}{100 / 18}=frac{36}{76} ) Hence( , Delta mathrm{p}=frac{36}{76} times 760=36 mathrm{mmHg} ) ( p=p_{1}^{circ}+Delta p=760-724 mathrm{mmHg} ) Regards
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