Upto first end point gm equi. of (NaOH+Na2CO3)=HCl
x+y×1=101×17.5
x+y=1.75....(1)
Upto second end point
NaOH+Na2CO3≡HCl
x+y×2=101×19
x+2y=1.9....(2)
y=0.15
=3.975
=4
Hence answer is (4)
0.4g mixture of NaOH, Na_2 CO_3 and some inert impurities was first titrated with N 10 HCl using phenolphthalein as an indicator, 17.5 mL of HCl was…
JEE Main 2021 — Chemistry Physical Chemistry
2021integereasy
0.4g mixture of NaOH,Na2CO3 and some inert impurities was first titrated with 10NHCl using phenolphthalein as an indicator, 17.5mL of HCl was required at the end point. After this methyl orange was added and titrated. 1.5mL of same HCl was required for the next end point. The weight percentage of Na2CO3 in the mixture is (Rounded-off to the nearest integer)
Official previous-year question
Held on 25 Feb 2021 · Verified 6 Jul 2026.
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Physical Chemistry in other years
More JEE Main Physical Chemistry questions
What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?
2026medium
  Which of the following point in Figure 2 most accurately represents the nodal surface as shown in Figure 1?
2026medium
At 298 K, the mole percentage of $\mathrm{N}_{2}(\mathrm{~g})$ in air is $80 \%$. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of $\mathrm{N}_{2}(\mathrm{~g})$ in water at 298 K ? ($\mathrm{K}_{\mathrm{H}}$ for $\mathrm{N}_{2}$ is $6.5 \times 10^{7} \mathrm{~mm} \mathrm{Hg}$)
2026medium
Electricity is passed through an acidic solution of $\mathrm{Cu}^{2+}$ till all the $\mathrm{Cu}^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is $\_\_\_\_$ mL. (Nearest integer) [Given: $\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) \mathrm{E}_{\text {red }}^{\mathrm{o}}=+0.34 \mathrm{~V}$ $\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \mathrm{E}_{\text {red }}^{\mathrm{o}}=+1.23 \mathrm{~V}$ Molar mass of $\mathrm{Cu}=63.54 \mathrm{~g} \mathrm{~mol}^{-1}$ Molar mass of $\mathrm{O}_{2}=32 \mathrm{~g} \mathrm{~mol}^{-1}$ Faraday Constant $=96500 \mathrm{C} \mathrm{mol}^{-1}$ Molar volume at $\mathrm{STP}=22.4 \mathrm{~L}$]
2026hard
Arrange the following resultant mixtures in increasing order of their pH values A. $10$ mL $0.2$ M Ca(OH)$_2$ + $25$ mL $0.1$ M HCl B. $10$ mL $0.01$ M H$_2$SO$_4$ + $10$ mL $0.01$ M Ca(OH)$_2$ C. $10$ mL $0.1$ M H$_2$SO$_4$ + $10$ mL $0.1$ M KOH Choose the correct answer from the options given below:
2026medium
Other Chemistry topics for JEE Main
Keep practicing Physical Chemistry
Work through every JEE Main Physical Chemistry PYQ, year by year.