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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?A. `32.25 mL`B. `45.98 mL`C. `78.60 mLD. `25.22 mL` |
Answer» Correct Answer - 4 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` |
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