CGPET 2019 Solved Paper

Given,
mass of Mg(W)=1.2g
molar mass of Mg(M)=24g
volume of 1M H2SO4=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)
=

n′×1000

V

=

0.05×1000

100

=0.5M
Hence, option (d) is the correct answer.