The volume of ethyl alcohol (density $1.15 g / c c$ ) that has to be added to prepare $100 c c$ of $0.5 M$ ethyl alcohol solution in water is :

1.15 c c

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2 c c

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2.15 c c

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2.30 c c

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Solution

The correct option is C $2 c c$
Molarity $= \frac{m a s s \times 1000}{m o l e c u l a r w e i g h t \times v o l u m e o f s o l u t i o n}$

Molecular weight of ethyl alcohol,
$C_{2} H_{5} O H = 12 \times 2 + 5 + 16 + 1$
$= 24 + 22 = 46 g {m o l}^{- 1}$

On substituting the values, we get
$0.5 M = \frac{m a s s \times 1000}{46 \times 100}$
Mass $= \frac{0.5 \times 46}{10} = 2.3 g$

$∴$ Volume of ethyl alcohol required,
$V = \frac{m a s s o f e t h y l a l c o h o l}{d e n s i t y o f e t h y l a l c o h o l} = \frac{2.3 g}{1.15 g / c c} = 2 c c$

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