Question

To dissolve 0.9 g metal, 100 mL of 1 N HCl is used. What is the equivalent weight of metal?

A 7
B 9
C 10
D 6
​

Answer 1

Answer:

0.9gms metal requires 100ml of 1N HCl

⇒ Volume of HCl=100ml

Normality of HCl=1N

⇒ Weight of HCl=

1000

N.Eq.Wt×V

[∵ Normality formula] =

1000

1×1×100

=0.1gm

We know, Equivalent Weight of Metal is the mass of metal required to produce 1gm of H

2

⟶(1)

Also, We know 0.1gm of HCl gives 0.1gm H

2

⇒ 1gm HCl Gives 1gm H

2

Now, if 0.9gm metal requires 0.1gm HCl

⇒1gm HCl will require

0.1

1×0.9

gm metal

∴ Mass of metal which produces 1gm H

2

=9gm

∴ From (1) Equivalent Weight of Metal= 9gm

Answer 2

Answer:

0.9gms metal requires 100ml of 1N HCl

⇒ Volume of HCl=100ml

Normality of HCl=1N

⇒ Weight of HCl=

1000

N.Eq.Wt×V

[∵ Normality formula] =

1000

1×1×100

=0.1gm

We know, Equivalent Weight of Metal is the mass of metal required to produce 1gm of H

2

⟶(1)

Also, We know 0.1gm of HCl gives 0.1gm H

2

⇒ 1gm HCl Gives 1gm H

2

Now, if 0.9gm metal requires 0.1gm HCl

⇒1gm HCl will require

0.1

1×0.9

gm metal

∴ Mass of metal which produces 1gm H

2

=9gm

∴ From (1) Equivalent Weight of Metal= 9gm