Molar mass of methanol \(\left(\mathrm{CH}_3 \mathrm{OH}\right)=32 \mathrm{~g} \mathrm{~mol}^{-1}\) \(=0.032 \mathrm{~kg} \mathrm{~mol}^{-1}\)
As Molarity
\(=\frac{\text { mass of solute }}{\text { molar mass of solute } \times \text { volume of solution }}\)
We can write molarity \(=\frac{\text { density of solution }}{\text { molar mass of solute }}\)
Thus the molarity of the given solution will be
\(\frac{0.793 \mathrm{~kg} \mathrm{~L}^{-1}}{0.032 \mathrm{~kg} \mathrm{~mol}^{-1}}=24.78 \mathrm{~mol} \mathrm{~L}^{-1}\)
Applying \(\mathrm{M}_1 \times \mathrm{V}_1=\mathrm{M}_2 \times \mathrm{V}_2\)
\(\begin{array}{ll}\text { (Given } & \text { (Solution to } \\ \text { solution) } & \text { be prepared) }\end{array}\)
\(24.78 \times \mathrm{V}_1=0.25 \times 2.5 \mathrm{~L}\)
or \(\mathrm{V}_1=0.02522 \mathrm{~L}=25.22 \mathrm{~mL}\)