Soln:
Volume of Naoh(V1)= 30cc
Normality of Naoh(N1)= 1N
Volume of excess H2so4(V2) = ??
Normality of H2so4(N2) = molarity × basicity
= 2×2 = 4 × 1.01 = 4.04N
Using,
V1N1 = V2N2
30×1 = V2 × 4.04
I.e. V2 = 7.42cc
Volume of excess acid = 7.42 cc
Volume of H2so4 neutralized by metal= 100 - 7.42
= 92.575 cc
Therefore, gm.eqv of H2so4 = gm.eqv of metal
(92.575/1000) × 4.04 = 4/E
I.e E = 10.09
Atomic mass= E × basicity
= 10.09 × 2
=20.18amu ans
