JEE MAIN - Chemistry (2008 - No. 26)
The vapour pressure of water at 20oC is 17.5 mm Hg. If 18 g of glucose (C6H12O6) is added to 178.2 g of water at 20oC, the vapour pressure of the resulting solution will be
17.675 mm Hg
15.750 mm Hg
16.500 mm Hg
17.325 mm Hg
Explanation
NOTE : On addition of glucose to water, vapour pressure of water will decrease. The vapour pressure of a solution of glucose in water can be calculated using the relation
$${{{P^ \circ } - {P_S}} \over {{P_S}}} = {{Moles\,\,of\,\,glu\cos e\,\,in\,\,solution} \over {moles\,\,of\,\,water\,\,in\,\,solution}}$$
or $$\,\,\,\,\,\,\,\,\,$$ $${{17.5 - {P_S}} \over {{P_S}}} = {{18/180} \over {178.2/18}}$$
[as $$\,\,\,\,\,\,\,\,\,$$ $${P^ \circ } = 17.5$$]
or $$\,\,\,\,\,\,\,\,\,$$ $$17.5 - {P_S} = {{0.1 \times {P_S}} \over {9.9}}$$
or $$\,\,\,\,\,\,\,\,\,$$ $${P_S} = 17.325\,mm\,Hg.$$
Hence (d) is correct answer.
$${{{P^ \circ } - {P_S}} \over {{P_S}}} = {{Moles\,\,of\,\,glu\cos e\,\,in\,\,solution} \over {moles\,\,of\,\,water\,\,in\,\,solution}}$$
or $$\,\,\,\,\,\,\,\,\,$$ $${{17.5 - {P_S}} \over {{P_S}}} = {{18/180} \over {178.2/18}}$$
[as $$\,\,\,\,\,\,\,\,\,$$ $${P^ \circ } = 17.5$$]
or $$\,\,\,\,\,\,\,\,\,$$ $$17.5 - {P_S} = {{0.1 \times {P_S}} \over {9.9}}$$
or $$\,\,\,\,\,\,\,\,\,$$ $${P_S} = 17.325\,mm\,Hg.$$
Hence (d) is correct answer.
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