Question

$29.2 \%( w / w ) HCl$ stock solution has a density of $1.25 g$ $mL ^{-1}$. The molecular weight of $HCl$ is $36.5 g mol ^{-1} .$ The volume (in $mL$ ) of stock solution required to prepare a $200 mL$ solution of $0.4 M HCl$ is-

• A. $16 \,mL$
• B. $61\, mL$
• C. $80 \,mL$
• D. $8 \,mL$

Answer 1

Answer: (D)

Density of solution $=1.25 \,g / mL$

$29.2 \%( w / w )$ means that $29.2 g$ of $HCl$ is present in 100 gms of solution

So, $\rho$ ( density of solution )$=\frac{\text { wt. of soln. }}{\text { vol. of soln. }}$

$1.25 g / mL =\frac{100}{\text { vol. of soln }}$

$v =\frac{100}{1.25} \,mL$

molarity of solution $=\frac{\text { no. of moles of solute }}{\text { vol. of solution (in ltrs) }}$

$M =\frac{29.2 / 36.5}{100 / 1.25} \times 1000=10 M$

Apply, $M _{1} V _{1}= M _{2} V _{2}$

$0.4 \times 200=10 \times V$

$V =8 \,mL$