Question

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?​

Answer 1

Volume of H₂O added = 175 ml

Further explanation

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 :

`\tt (45+x)/(100+x)=80\%\\\\45+x=0.8(100+x)\\\\45+x=80+0.8x\\\\35=0.2x\rightarrow x=175~g`

For water mass=volume(density = 1 g/ml)

So volume added = 175 ml