x mg of pure HCl was used to make an aqueous solution. 25.0 mL of 0.1 M Ba(OH)2 solution is used when the HCl solution was titrated against it. The numerical value of x is ____ × 10−1. (Nearest integer) Given : Molar mass of HCl and Ba(OH)2 are 36.5 and 171.0 g mol−1 respectively.

x mg of pure HCl was used to make an aqueous solution

Q. x mg of pure HCl was used to make an aqueous solution. 25.0 mL of 0.1 M Ba(OH)₂ solution is used when the HCl solution was titrated against it. The numerical value of x is ____ × 10⁻¹. (Nearest integer)

Given : Molar mass of HCl and Ba(OH)₂ are 36.5 and 171.0 g mol⁻¹ respectively.

Correct Answer: 1825

Explanation

Balanced chemical equation:

Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O

Moles of Ba(OH)₂ used:

$$ = M \times V = 0.1 \times \frac{25}{1000} = 2.5 \times 10^{-3}\ \text{mol} $$

From stoichiometry:

1 mol Ba(OH)₂ reacts with 2 mol HCl

Moles of HCl used:

$$ = 2 \times 2.5 \times 10^{-3} = 5.0 \times 10^{-3}\ \text{mol} $$

Mass of HCl:

$$ = n \times M = 5.0 \times 10^{-3} \times 36.5 = 0.1825\ \text{g} $$

Convert into mg:

$$ 0.1825\ \text{g} = 182.5\ \text{mg} $$

Given format:

$$ x \times 10^{-1} = 182.5 $$

$$ x = 1825 $$

Hence, the required value of x is 1825.

Explanation

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