An aqueous solution is prepared by diluting 3.30 mL of acetone (density = 0.789 g/mL) to get the final volume of 75 mL. Find the mole fraction of acetone. (Assume no volume change on mixing).

0.11

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0.1

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0.001

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0.01

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Solution

The correct option is D 0.01
We know that:
$D e n s i t y = \frac{M a s s}{V o l u m e}$
$M a s s o f a c e t o n e = v o l u m e \times d e n s i t y = 3.30 m L \times 0.789 g m L^{- 1} = 2.603 g$

Number of moles of acetone in the solution $= \frac{2.603 g}{58 g m o l^{- 1}} = 0.044 m o l$
$M a s s o f w a t e r = (F i n a l v o l u m e - i n i t i a l v o l u m e) \times d e n s i t y o f w a t e r$

We know that density of water is 1.00 g/mL

$M a s s o f w a t e r = (75.00 m L - 3.30 m L) \times 1 g m L^{- 1} = 71.70 g$

Number of moles of water $= \frac{71.7}{18.0} = 3.98 m o l$

We know that:
$Molefraction=number of moles of speciestotal number of moles$

$Mole fraction of acetone=0.0440.044+3.98=0.010$

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