If the density of methanol is 0.793 kg` L^(–1)`, what is 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)`
Molarity of the given solution
`=(W_(2) "in" kg)/(Mw_(2)xxV_(sol)(L))=(d_(sol)(kgL^(-1)))/(Mw_(2)(kg))`
`=(0.793 kg L^(-1))/(0.032 kg mol^(-1))=24.78 M`
Applying `underset(("Given solution"))(M_(1)xxV_(1))=underset("prepared)")underset("(Solution to be")(M_(2)V_(2)`
`24.78xxV_(1)=0.25xx2.5 L`
or `V_(1)=0.02522 L=25.22 mL`