Molar mass of methanol `(CH_(3)OH)= (1xx12)+(4xx1)+(1xx16)`
`=32 g mol ^(-1)`
`=0.032 kg mol^(-1)`
Molarity of methanol solution `=(0.793 kg l^(-1))/(0.032 kg mol^(-1))`
`= 24.78 mol L^(-1)`
(Since density is mass per unit volume ) Applying ,
`M_(1)V_(1)=M_(2)V_(2)`
(Given solution )( slution to be prepared ) (24.78 mol `L^(-1)`) `V_(1)=(2.5 L) (0.25 mol L^(-1))`
`V_(1)= 0.02252 L`
`V_(1) =25.22 mL`