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Q.
A mixture has $ 18 \, g $ water and $ 414\,g $ ethanol. The mole fraction of water in mixture is (assume ideal behaviour of the mixture)
Haryana PMTHaryana PMT 2008
Solution:
Mole fraction $=\frac{\text { moles of water }}{\text { moles of water }+\text { moles of ethanol }}$
Mol. wt. of $H _{2} O =18 \,mol$.
wt. of $C _{2} H _{5} OH =46$
Mole fraction of water $=\frac{18 / 18}{18 / 18+414 / 46}$
$=\frac{1}{1+9}=\frac{1}{10}=0.1$