1.

A solution contains 25% water, 25% ethanol (C_(2)H_(5)OH ) and 50% acetic acid (CH_(3)COOH) by mass. The mole fraction of

Answer»

water =0.502
ETHANOL = 0.302
acetic ACID = 0.196
ethanol + acetic acid = 0.497

Solution :MOLES of water `=25/18 = 1.388`
Moles of ethanol `=25/46 = 0.543`
Moles of acetic acid `=50/60 = 0.833`
TOTAL moles `=1.388 + 0.543 + 0.833 = 2.764`
`x_("water") = 1.388/2.764 = 0.502`
`e_("ethanol") = 0.196`
`e_("acetic acid") = 0.301`
`x_("ethanol") + x_("ACETI acid") = 0.497`


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