If the density of methanol is `0.793 kg L^(-1)` what ia its volume needed for making 2.5 L of its `0.25 M` solution?

Explanation

`"Molar mass of methanol "(CH_(3)OH)="32 g mol"^(-1)="0.032 kg mol"^(-1)`
`{:("Applying "," "M_(1)xxV_(1),=," "M_(2)V_(2)),(," (Given solution)",,"(Solution to be prepared)"),(,24.78xxV_(1)=0.25xx2.5L,,):}`
`"or "V_(1)=0.02522L=25.22mL`