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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 glycerin is :

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a

0.40

b

0.46

c

0.36

d

0.20

answer is C.

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Detailed Solution

Mole of glycerin,

$X = \frac{M_{G}}{M_{W} + M_{G}}$

$X = \frac{\frac{46}{92}}{\frac{46}{92} + \frac{36}{18}}$

$X = \frac{0.5}{0.5 + 2} X = 0.20$

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