Question

17) To neutralize 25 ml of o.25 N Na2CO3 solution how much volume of

0.5 N HCl required ?

A) 30 ml

B) 25 ml

C) 0.50 ml

D) 0.53 ml​

Answer 1

Answer:

25 ml

Explanation:

By law of equivalence:

N1 V 1=N 2 V2

2×0.25M×25=1×0.5M×V 2

V 2= 2×0.25M×25

0.5m

=25ml