Density of methanol is `0.793kg*L^(-1)`. What is its volume needed for making 2.5L of its 0.25 (M) solution?

Explanation

Molecular mass of methanol `(CH_(3)OH)=32`
No. of moles of methanol present in 1L 0.25 (M) solution=0.25
`therefore` Number of moles of methanol present in 2.5 L 0.25 (M)
solution `=0.25xx2.5=0.625`
Mass of 0.625 mol `CH_(3)OH=0.625xx32=20g`
Density of `CH_(3)OH=0.793kg*L^(-1)`
or, `0.793xx10^(3)g*L^(-1)`
Hence, required volume of methanol
`=("mass of methanol")/("density of methanol")=(20g)/(0.793xx10^(3)g*L^(-1))`
=0.02522L=25.22mL
`therefore 25.22mL` of methanol is required.