Question

1
2.08g of zinc displaced 6.9119g of silver from a solution of silver nitrate. If the equivalent weight of
silver is 108, then the equivalent weight of zinc is
a) 65.00
b) 32.5
c) 64.50
d) 31.50​

Answer 1

By law of equivalence, the no. of gram equivalence of each reactant should be same, i.e.,

$\tt{\longrightarrow e_1=e_2}$

$\tt{\longrightarrow \dfrac{W_1}{E_1}=\dfrac{W_2}{E_2}}$

$\tt{\longrightarrow E_2=\dfrac{W_2E_1}{W_1}}$

In this question,

• given mass of silver, $\tt{W_1=6.9119\ g}$

• given mass of zinc, $\tt{W_2=2.08\ g}$

• equivalent weight of silver, $\tt{E_1=108\ g\,eq^{-1}}$

Then equivalent weight of zinc will be,

$\tt{\longrightarrow E_2=\dfrac{W_2E_1}{W_1}}$

$\tt{\longrightarrow E_2=\dfrac{2.08\times108}{6.9119}}$

$\tt{\longrightarrow\underline{\underline{E_2=32.5\ g\,eq^{-1}}}}$

Hence (b) is the answer.

Answer 2

By law of equivalence, the no. of gram equivalence of each reactant should be same, i.e.,

\tt{\longrightarrow e_1=e_2}⟶e

1

=e

2

\tt{\longrightarrow \dfrac{W_1}{E_1}=\dfrac{W_2}{E_2}}⟶

E

1

W

1

=

E

2

W

2

\tt{\longrightarrow E_2=\dfrac{W_2E_1}{W_1}}⟶E

2

=

W

1

W

2

E

1

In this question,

given mass of silver, \tt{W_1=6.9119\ g}W

1

=6.9119 g

given mass of zinc, \tt{W_2=2.08\ g}W

2

=2.08 g

equivalent weight of silver, \tt{E_1=108\ g\,eq^{-1}}E

1

=108 geq

−1

Then equivalent weight of zinc will be,

\tt{\longrightarrow E_2=\dfrac{W_2E_1}{W_1}}⟶E

2

=

W

1

W

2

E

1

\tt{\longrightarrow E_2=\dfrac{2.08\times108}{6.9119}}⟶E

2

=

6.9119

2.08×108

\tt{\longrightarrow\underline{\underline{E_2=32.5\ g\,eq^{-1}}}}⟶

E

2

=32.5 geq

−1

Hence (b) is the answer.