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?

Options

A. `32.25 mL`
B. `45.98 mL`
C. `78.60 mL
D. `25.22 mL`

Answer

4

Explanation

It is basically a problem of dilution of solution, which can be solved by applying the molarity equation:
`M_(i) V_(i) =M_(f)V_(f)`
our objective is to get `V_(i)`. This is possible provided we know `M_(i)`.
By definition, `M=n_(solute)/V_(L)`
Since `n_(solute)=(mass_(solute))/(molar mass_(solute))`
We can write `M=(mass_(solute))/(V_(L)xxmolar mass_(solute))`
Since in this problem, we need to dilute `CH_(3)OH, V_(L)` is equal to volume of `CH_(3)OH` in litres and mass/volume is density. Thus
`M=((density)_(CH_(3)OH)" in "Kg L^(-1))/((molar mass_(CH_(3)OH))" in "Kg mol^(-1))`
`=(0.793 Kg L^(-1))/(0.032 Kg mol^(-1))`
`=24.78 mol L^(-1)=M_(i)`
Now `24.78 mol L^(-1) xxV_(i)=0.25 mol L^(-1)xx2.5 L`
Thus, `V_(i)=((0.25 mol L^(-1))(2.5 L))/((23.78 mol L^(-1)))=0.02522 L`
`=25.22 mL`