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Raoult's Law

Determine mol...

Question

Determine mole fraction of $C H_{3} O H$ in an aqueous solution of $C H_{3} O H$ whose molality is 2.00 m.

0.35

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0.035

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0.65

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0.965

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Solution

The correct option is B 0.035
2.00 m means 2 moles of $C H_{3} O H$ is dissolved in 1 kg of water.
Number of moles of water in 2.00 m of solution $= \frac{1000 g}{18 g m o l^{- 1}} = 55.55 m o l$
Number of moles of $C H_{3} O H$ present in 2.00 m solution = 2 mol

Mole fraction of $C H_{3} O H$
$= \frac{m o l e s o f C H_{3} O H i n s o l u t i o n}{t o t a l n u m b e r o f m o l e s i n s o l u t i o n} = \frac{2}{(2 + 55.55)} = 0.035$

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