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,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)
Held on 25 Feb 2021 · Verified 6 Jul 2026.
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