Question.
If the density of methanol is $0.793 \mathrm{~kg} \mathrm{~L}^{-1}$, what is its volume needed for making $2.5 \mathrm{~L}$ of
its 0.25 M solution?
If the density of methanol is $0.793 \mathrm{~kg} \mathrm{~L}^{-1}$, what is its volume needed for making $2.5 \mathrm{~L}$ of
its 0.25 M solution?
Solution:
Molar mass of methanol $\left(\mathrm{CH}_{3} \mathrm{OH}\right)=(1 \times 12)+(4 \times 1)+(1 \times 16)$
$=32 \mathrm{~g} \mathrm{~mol}^{-1}$
$=0.032 \mathrm{~kg} \mathrm{~mol}^{-1}$
Molarity of methanol solution $=\frac{0.793 \mathrm{~kg} \mathrm{~L}^{-1}}{0.032 \mathrm{~kg} \mathrm{~mol}^{-1}}$
$=24.78 \mathrm{~mol} \mathrm{~L}^{-1}$
(Since density is mass per unit volume)
Applying,
$\mathrm{M}_{1} \mathrm{~V}_{1}=\mathrm{M}_{2} \mathrm{~V}_{2}$
(Given solution) (Solution to be prepared)
$\left(24.78 \mathrm{~mol} \mathrm{~L}^{-1}\right) \mathrm{V}_{1}=(2.5 \mathrm{~L})\left(0.25 \mathrm{~mol} \mathrm{~L}^{-1}\right)$
$\mathrm{V}_{1}=0.0252 \mathrm{~L}$
$\mathrm{V}_{1}=25.22 \mathrm{~mL}$
Molar mass of methanol $\left(\mathrm{CH}_{3} \mathrm{OH}\right)=(1 \times 12)+(4 \times 1)+(1 \times 16)$
$=32 \mathrm{~g} \mathrm{~mol}^{-1}$
$=0.032 \mathrm{~kg} \mathrm{~mol}^{-1}$
Molarity of methanol solution $=\frac{0.793 \mathrm{~kg} \mathrm{~L}^{-1}}{0.032 \mathrm{~kg} \mathrm{~mol}^{-1}}$
$=24.78 \mathrm{~mol} \mathrm{~L}^{-1}$
(Since density is mass per unit volume)
Applying,
$\mathrm{M}_{1} \mathrm{~V}_{1}=\mathrm{M}_{2} \mathrm{~V}_{2}$
(Given solution) (Solution to be prepared)
$\left(24.78 \mathrm{~mol} \mathrm{~L}^{-1}\right) \mathrm{V}_{1}=(2.5 \mathrm{~L})\left(0.25 \mathrm{~mol} \mathrm{~L}^{-1}\right)$
$\mathrm{V}_{1}=0.0252 \mathrm{~L}$
$\mathrm{V}_{1}=25.22 \mathrm{~mL}$