JEE Main 2021 — Some Basic Concepts Of Chemistry Question with Solution
From: JEE Main 2021 (Online) 18th March Evening Shift
Question
10.0 mL of Na2CO3 solution is titrated against 0.2 M HCl solution. The following titre values were obtained in 5 readings :
4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL and 5.0 mL
Based on these readings, and convention of titrimetric estimation the concentration of Na2CO3 solution is ___________ mM.
(Round off to the Nearest Integer).
4.8 mL, 4.9 mL, 5.0 mL, 5.0 mL and 5.0 mL
Based on these readings, and convention of titrimetric estimation the concentration of Na2CO3 solution is ___________ mM.
(Round off to the Nearest Integer).
Enter your answer
Show full solutionCorrect answer: 50
Correct answer
50
Step-by-step explanation
From the given value of HCl, it is clear that most appropriate volume of HCl used is 5 ml because it occurs most number of times.
Na2CO3 + 2 HCl 2 NaCl + CO2 + H2O
n factor of Na2CO3 = 2
n factor of HCl = 1
equivalent of Na2CO3 = equivalent of HCl
= 0.05 M
= 50 103 M
= 50 mM
Na2CO3 + 2 HCl 2 NaCl + CO2 + H2O
n factor of Na2CO3 = 2
n factor of HCl = 1
equivalent of Na2CO3 = equivalent of HCl
= 0.05 M
= 50 103 M
= 50 mM
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