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 mol−1. The volume (mL) of stock solution required to prepare a 200 mL solution 0.4 M HCl is:

• A. $6ml$
• B. $8ml$
• C. $10ml$
• D. $12ml$

Answer 1

The correct option is B

$8ml$

Explanation for the correct answer:-

Option (B)$8ml$

Step 1: Given data :

Density of $29.2\%$(w/W) $\mathrm{HCl}$ stock solution = $1.25gm{L}^{-1}$

Molecular weight of$\mathrm{HCl}$ = $36.5mo{l}^{-1}$

Step 2: Formula used :

Molarity = $\frac{Givenmass}{molarmass}\times \frac{1000}{vol}$

Step 3: Calculation of Molarity

The molarity of stock solution of$\mathrm{HCl}$

$$\begin{gathered} =\frac{29.2\times 1.25\times 10}{36.5} \\ =\frac{365}{36.5} \\ =10M \end{gathered}$$

Step 4: Applying the dilution equation

$$\begin{gathered} M_i{V}_i=M_f{V}_f \\ {V}_i=\frac{0.4\times 200}{10} \\ =\frac{80}{10} \\ =8ml \end{gathered}$$

Hence option B is correct.