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

Let us calculate moles of methanol present in 2.5 L of 0.25 M solution.
Molarity `=("Moles of" CH_(3)OH)/("Volume in L")`
`0.25 = ("Moles of" CH_(3)OH)/2.5`
`therefore` Moles of `CH_(3)OH = 0.25 xx 2.5 = 0.625` moles
Mass of `CH_(3)OH = 0.625 xx 32 = 20`g
(Molecular mass of `CH_(3)OH=32`)
Now `0.793 xx 10^(3)`g of `CH_(3)OH` is present in 1000 mL
20 g of `CH_(3)OH` will be present in `=(1000)/(9.793 xx 10^(3)) xx 20 = 25.2` mL