Volume of H₂O added = 175 ml
<h3>Further explanation</h3>
Given
100 gm of a 55% (M/M) and 20% (M/M) nitric acid solution
Required
waters added
Solution
starting solution
mass H₂O = 45%=45 g
%mass of H₂O in new solution = 100%-20%=80%
Can be formulated for %mass H₂O :

For water mass=volume(density = 1 g/ml)
So volume added = 175 ml
Answer:
A(g)+B(g)⟶C(g) 2A(g)+2B(g)⟶5C(g)
A(s)+B(s)⟶C(g) 2A(g)+2B(g)⟶3C(g):