TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Reaction of neutralisation
NaOH+H2SO4⟶Na2SO4+H2O
Number of equivalent of NaOH=
Number of equivalent of H2SO4
Number of equivalent mol‌H2SO4=100×0.5
=50mmol
50×n-factor ⇒50×2=100 milli equivalent.
Number of Eq. of NaOH
‌‌=N×100mL(N=‌ Normality ‌)
‌‌=100N
Number of Eq. of NaOH=
Number of Eq. of H2SO4
100N‌‌=100
N‌‌=1
Number of mole of NaOH in 1000mL=1
Number of mole of NaOH in 100mL=0.1
Mass of NaOH in 100mL=0.1×40=4g
Mass of Na2SO4=100−4=96g