JEE Main Chemistry Class 12 Analytical Chemistry Questions

As given, NaOH and Na2CO3 is titrated with N∕10HCl.
For NaOH,
Equivalents of NaOH=
Equivalents of HCl
Equivalents of HCl= Normality × Volume (L)
=0.1×‌

17.5

1000

Equivalent of HCl=1.75×10−3
Equivalents of NaOH=1.75×10−3
Weight of NaOH= Equivalent of NaOH× equivalent weight of
=40×1.75×10−3=0.07g
NaOH
Now, weight % of NaOH
=‌

0.07

0.4

×100=‌

70

4

=17.5%
Similarly for Na2CO3,
Equivalent of HCl= Equivalent of Na2CO3
Equivalent of Na2CO3=0.1×‌

1.5

1000

=0.15×10−3
Weight of Na2CO3= Equivalent of Na2CO3 × equivalent weight of Na2CO3
=0.15×10−3×106=15.9×10−3g
Weight \% of Na2CO3=‌

15.9×10−3

0.4

×100
=0.039×100=3.9%∼eq4%