Given, mass of Mg(W)=1.2g molar mass of Mg(M)=24g volume of 1MH2SO4=100mL Thus, ∵ moles (n) of Mg=
W
M
=
1.2
24
=0.05‌mole and moles (n) of H2SO4 in 100mL solution of 1MH2SO4=M×V(L) =
1×100
1000
=0.1 moles. Where, (V= volume of solution) Therefore, H2SO4 left after treated with 100mL of 1MH2SO4=0.1−0.05 i.e., n′=0.05 moles Thus, concentration of H2SO4 (left) =