100gm of a 55% (M/M) nitric acid solution is to be diluted to 20% (M/M) nitric acid. How many liters of water should be added to the starting solution?

Volume of H₂O added = 175 ml

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

these nuts

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2