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  4. Mole fraction of C3H5(OH)3 in a solution of 36 g of water and 46 g of glycerine is:

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Q. Mole fraction of $C_{3}H_{5}(OH)_{3}$ in a solution of 36 g of water and 46 g of glycerine is:

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Solution:

Mole of $H_{2}O=\frac{36}{18}=2$
Mole of glycerine $=\frac{46}{92}=0.5$
Total moles $= 2 + 0.5 = 2.5$
Mole fractions of glycerine $=\frac{n_{1}}{n_{1}+n_{2}}=\frac{0.5}{2.5}$
$=0.2$