Question

What is the volume of 0.1 N HCl required to react completely with 1.0 g of pure calcium carbonate​

Answer 1

Answer:

200cm3

Explanation:

N=0.1, w=1.0gm, Equivalent weight CaCO

3

=50, V

HCl

=?

Using the law of equivalence:

n

eq

=N×V⇒V=

N

n

eq

V=

Eq.wt.×N

w×1000

=

50×0.1

1×1000

=200cm

3

Answer 2

Answer:

N = 0.1

Normality, N = W ×1000/ M × V

Where, W = equivalent weight of CaCO3=100/2 = 50

M = Molar mass of CaCO3 = 100

V = Volume of HCl required =V

Therefore,

0.1 = 1 ×1000/ 50× V

V = 200ml

hope this will help you.