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Q.
A solution contains $ 25%\text{ }Hp,\text{ }25%\,{{C}_{2}}{{H}_{5}}OH $ and 50% $ C{{H}_{3}}COOH $ by mass. The mole fraction of $ {{H}_{2}}O $ would be
MGIMS WardhaMGIMS Wardha 2010
Solution:
Mole fraction of $ {{H}_{2}}O $ $ =\frac{number\text{ }of\text{ }moles\text{ }of}{total\text{ }number\text{ }of\text{ }moles\text{ }of\text{ }all\text{ }components} $ Let the total mass of solution $ =100\text{ }g $ Mass of $ {{H}_{2}}O=25g $ Mass of $ {{C}_{2}}{{H}_{5}}OH=25g $ Mass of $ C{{H}_{3}}COOH=50\text{ }g $ Moles of $ {{H}_{2}}O=\frac{25}{18}=1.388 $ ( $ \because $ Molar mass of $ {{H}_{2}}O=18 $ ) Moles of $ {{C}_{2}}{{H}_{5}}OH=\frac{25}{46}=0.543 $ ( $ \because $ Molar mass $ {{C}_{2}}{{H}_{5}}OH=46 $ ) Moles of $ C{{H}_{3}}COOH=\frac{50}{60}=0.833 $ ( $ \because $ Molar mass of $ C{{H}_{3}}COOH=60 $ ) Total no. of moles $ =1.388+0.543+0.833 $ $ =2.764 $ $ \therefore $ Mole fraction of $ {{H}_{2}}O=\frac{1.388}{2.764} $ $ =0.502 $