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 mLI 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

Calculate the volume (mL) of concentrated acid required to prepare 500

1.	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 mLI 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» 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.

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