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  4. If the density of methanol is 0.793 kg L-1, what volume of it is needed for making 2.5 L of its 0.25 M solution?

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Q. If the density of methanol is $0.793 \,kg \,L^{-1}$, what volume of it is needed for making $2.5\, L$ of its $0.25\, M$ solution?

Some Basic Concepts of Chemistry

Solution:

Molarity $= \frac{\text{Moles of}\,CH_3OH}{\text{Volume in}\,L}$
$\Rightarrow 0.25 = \frac{\text{moles of }CH_3OH}{2.5}$
$\therefore $ Moles of $CH_3OH = 2.5 \times 0.25 = 0.625$
Mass of $CH_3OH = 0.625 \times 32 = 20 \,g$
[Molar mass of $CH_3OH = 12 + 1 \times 3 + 16 + 1 = 32]$
$0.793 \times 10^3 \,g$ of $CH_3OH$ is present in $1000 \,mL$.
$\therefore 20\,g $ of $CH_3OH$ is present in
$ = \frac{1000}{0.793 \times 10^3} \times 20 = 25.2\,mL$