Verified Solution • 2025–26 Session
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QJEEM21Ch75af: JEE Main 2021 Chemistry - Class 11 JEEM21Ch75af Solution: 10.0 mL of Na 2 CO 3 solution is titrated against 0.2 M HCl…
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).
Step-by-Step Verified Solution
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
n factor of Na2CO3 = 2
n factor of HCl = 1
equivalent of Na2CO3 = equivalent of HCl
= 0.05 M
= 50
= 50 mM
Na2CO3 + 2 HCl
n factor of Na2CO3 = 2
n factor of HCl = 1
equivalent of Na2CO3 = equivalent of HCl
= 0.05 M
= 50
= 50 mM