Answered

Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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?​

Sagot :

ardni313

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 :

[tex]\tt \dfrac{45+x}{100+x}=80\%\\\\45+x=0.8(100+x)\\\\45+x=80+0.8x\\\\35=0.2x\rightarrow x=175~g[/tex]

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

So volume added = 175 ml

We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.