If the density of ${\mathrm{CH}}_3\mathrm{OH}$ is $0.80\mathrm{kg} {\mathrm{L}}^{-1}$, the volume of methanol to prepare $2.5\mathrm{L}$of $0.25\mathrm{M}$  aqueous solution is 

Since molarity, $M=n_2/V$ we have 

Amount of ${\mathrm{CH}}_3\mathrm{OH}$ required, $n_2=MV=\left(0.25{\mathrm{molL}}^{-1}\right)\left(2.5\mathrm{L}\right)=0.625\mathrm{mol}$

Mass of ${\mathrm{CH}}_3\mathrm{OH}$ required $m_2=n_2{M}_2=\left(0.625\mathrm{mol}\right)\left(32\mathrm{g}{\mathrm{mol}}^{-1}\right)=20.0\mathrm{g}=20.0\times {10}^{-3}\mathrm{kg}$

Volume of ${\mathrm{CH}}_3\mathrm{OH}$ required, $V=\frac{m_2}{\rho }=\frac{20.0\times {10}^{-3}\mathrm{kg}}{0.80{\mathrm{kgL}}^{-1}}=0.025\mathrm{L}=25.0\mathrm{mL}$.