If the density of methanol is $0.793 \mathrm{~kg} \mathrm{~L}^{-1}$,

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)

$\left(24.78 \mathrm{~mol} \mathrm{~L}^{-1}\right) \mathrm{V}_{1}=(2.5 \mathrm{~L})\left(0.25 \mathrm{~mol} \mathrm{~L}^{-1}\right)$

$\mathrm{V}_{1}=0.0252 \mathrm{~L}$

$\mathrm{V}_{1}=25.22 \mathrm{~mL}$