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  4. What will be the mole fraction of ethanol in a sample of spirit containing 85% ethanol by mass?

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Q. What will be the mole fraction of ethanol in a sample of spirit containing 85% ethanol by mass?

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Solution:

$x_{C_2H_5OH} = \frac{^{n}C_{2}H_{5}OH}{^{n}C_{2}H_{5}OH+\,{}^{n}H_{2}O}$
Mass of $C_{2}H_{5}OH = 85\, g$
Molar mass of $C_{2}H_{5}OH = 46 \,g/mol$
$^{n}C_{2}H_{5}OH= \frac{85}{46}=1.85\,mol$
Mass of water $= 100 - 85 = 15 \,g$
$^{n}H_{2}O = \frac{15}{18} = 0.833\,mol$
$x_{C_2H_5OH} = \frac{1.85}{1.85+0.833}$
$= \frac{1.85}{2.683} = 0.69$