Question

Calculate the volume (mL) of concentrated acid required to prepare 500 mL of 0.25 N HCl solution from concentrated stock HCl solution (specific gravity = 1.19) and 37.2% (by mass).

(a) 12.128 mL

(b) 20.613 mL

(c) 10.307 mL

(d) 24.256 mL

I have been asked this question in an interview for internship.

The query is from Expressing Concentration of Solutions in section Solutions of Chemistry – Class 12

Answer 1

Correct answer is (c) 10.307 mL

Explanation: Given,

Final concentration, N2 = 0.25 N

Final volume, V2 = 500 mL = 0.5 L

Consider 100 kg of solution.

100 kg of stock solution contains 37.2 kg acid.

Volume of stock solution containing 37.2 kg acid = 100kg/(1.19 kg/L) = 84.0336 L

Number of moles of HCl acid, n = mass of acid/molar mass of acid

n = 37.2 kg/(1 + 35.5) = 1.01918 kmole = 1019.18 mole

Molarity, M = n/volume of stock solution = 1019.18/84.0336 = 12.1282 M

Normality of stock solution, N1 = nf x Molarity = 1 x 12.1282 = 12.1282 N

Using equivalence equation N1 x V1 = N2 x V2:

Volume of stock solution to be added, V1 = (0.25 x 500 mL)/(12.1282) = 10.307 mL.