Question

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

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

Answer 1

The Correct Option is C

Given, density of $CH _{3} OH =0.8\, kg\, L ^{-1}$
Molarity $=0.25\, M$. Volume of $0.25\, M =4 L$
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 $CH _{3} OH =12+3+16+1=32\, mol ^{-1}$
$\therefore 0.25\, mol ^{-1}=\frac{\text { Given mass }}{32\, g\,mol ^{-1}} \times \frac{1}{4 L}$
$\therefore$ Given mass $=32\, g$ or $0.032\, kg$
Again $\because$ denstiy
$=\frac{\text { given mass }( kg )}{\text { volume }( mL )}$
$\therefore 0.8\, kg\, L^{-1}=\frac{0.032\, kg }{V( mL )}$
or $0.8 \times 1000 kg\, mL ^{-1}=\frac{0.032\, kg }{V( mL )}$
$\therefore V(m L)=\frac{0.032\, kg }{0.8\, kg L^{-1}} \times 1000\, V (m L)=40$