Question

If the density of methanol is $0.8kg{L}^{-1}$, what is its volume needed for making $4L$ of its $0.25M$ solution?

• A. 4 mL
• B. 8 mL
• C. 40 mL
• D. 80 mL

Answer 1

Answer: (C)

Given, density of $C{H}_{3}OH=0.8kg{L}^{-1}$
Molarity $=0.25M$. Volume of $0.25M=4L$
Volume needed $=?$
First of all we find mass of methanol (i.e. given mass)
Molarity $=\frac{\text{ Number of moles }}{\text{ Volumeof solution(L) }}$
Molarity $=\frac{\text{ Given mass }}{\text{ Molar mass }}\times \frac{1}{\text{ Volume of solution(L) }}$
Molar mass of $C{H}_{3}OH=12+3+16+1=32mo{l}^{-1}$
$\therefore 0.25mo{l}^{-1}=\frac{\text{ Given mass }}{32gmo{l}^{-1}}\times \frac{1}{4L}$
$\therefore$ Given mass $=32g$ or $0.032kg$
Again $\because$ denstiy
$=\frac{\text{ given mass }(kg)}{\text{ volume }(mL)}$
$\therefore 0.8kg{L}^{-1}=\frac{0.032kg}{V(mL)}$
or $0.8\times 1000kgm{L}^{-1}=\frac{0.032kg}{V(mL)}$
$\therefore V(mL)=\frac{0.032kg}{0.8kg{L}^{-1}}\times 1000V(mL)=40$