Question

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

• A. 1.15 cc
• B. 2 cc
• C. 2.15 cc
• D. 2.30 cc

Answer 1

Answer: (B)

2 cc

Answer 2

Molarity $=\frac{\text { mass } \times 1000}{\text { molecular weight } \times \text { volume of solution }}$
Molecular weight of ethyl alcohol,
$C _{2} H _{5} OH =12 \times 2+5+16+1$
$=24+22=46 \,g \,mol ^{-1}$
On substituting the values, we get
$0.5\, M =\frac{\text { mass } \times 1000}{46 \times 100}$
Mass $=\frac{0.5 \times 46}{10}=2.3\, g$
$\therefore$ Volume of ethyl alcohol required,
$V=\frac{\text { mass of ethyl alcohol }}{\text { density of ethyl alcohol }}=\frac{2.3\, g }{1.15\, g / cc }=2\, cc$