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

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Some Basic Concepts of Chemistry

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 (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is

Answer: 8

Solution

First, calculate the molarity of the stock solution:
Molarity =

\frac{mass percent \times density \times 1000}{molecular weight \times 100}

Molarity =

\frac{29.2 \times 1.25 \times 1000}{36.5 \times 100} = \frac{36500}{3650} = 10

M

Using the dilution formula

M_{1} V_{1} = M_{2} V_{2}

10 \times V_{1} = 0.4 \times 200

V_{1} = \frac{80}{10} = 8

mL

Therefore, 8 mL of stock solution is required.

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Integer answer

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