JEE Advanced · Chemistry · 1. Mole Concept
\(29.2 \%(\mathrm{w} / \mathrm{w}) \mathrm{HCl}\) stock solution has a density of \(1.25 \mathrm{gmL}^{-1}\). The molecular weight of \(\mathrm{HCl}\) is \(36.5 \mathrm{~g} \mathrm{~mol}^{-1}\). The volume \((\mathrm{mL})\) of stock solution required to prepare a 200 \(\mathrm{mL}\) solution of \(0.4 \mathrm{M} \mathrm{HCl}\) is :
- A 2
- B 4
- C 6
- D 8
Answer & Solution
Correct Answer
(D) 8
Step-by-step Solution
Detailed explanation
Molarity of stock solution of \(\mathrm{HCl}\)
\(=\frac{29.2 \times 1000 \times 1.25}{100 \times 36.5}\)
Let the volume of stock solution required \(=V \mathrm{~mL}\)
Thus, \(V \times \frac{29.2 \times 1000 \times 1.25}{100 \times 36.5}=200 \times 0.4\) \(\Rightarrow V=8 \mathrm{~mL}\)
\(=\frac{29.2 \times 1000 \times 1.25}{100 \times 36.5}\)
Let the volume of stock solution required \(=V \mathrm{~mL}\)
Thus, \(V \times \frac{29.2 \times 1000 \times 1.25}{100 \times 36.5}=200 \times 0.4\) \(\Rightarrow V=8 \mathrm{~mL}\)
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