JEE MAIN - Chemistry (2006 - No. 30)
18 g of glucose (C6H12O6) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100oC is
759.00 Torr
7.60 Torr
76.00 Torr
752.40 Torr
Explanation
Moles of glucose $$ = {{18} \over {180}} = 0.1$$
Moles of water $$ = {{178.2} \over {18}} = 9.9$$
Total moles $$ = 0.1 + 9.9 = 10$$
$${P_{{H_2}O}} = $$ Mole fraction $$ \times $$ Total pressure
$$ = {{9.9} \over {10}} \times 760$$
$$ = 752.4\,\,$$ Torr
Moles of water $$ = {{178.2} \over {18}} = 9.9$$
Total moles $$ = 0.1 + 9.9 = 10$$
$${P_{{H_2}O}} = $$ Mole fraction $$ \times $$ Total pressure
$$ = {{9.9} \over {10}} \times 760$$
$$ = 752.4\,\,$$ Torr
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