A 1N solution of NaOH is diluted 4/20 then rediluted 1:50. What is the final normality? How many grams of NaOH are present in 100ml of the final solution?

Question

I sort of understand the basic concept, but I don't really know how to do normality problems.

Answer 1

This isn't so much a normality problem as it is a dilution (concentration) problem.
If you take 4 ml and dilute to 20 you have diluted it 5 times, right? Then another 50 times in the 1:50. So the total dilution is 5*50 = 250 times. So if you have 1N solution and it's diluted 250 times it is not 1/250 = 0.004N, right? But you can do it with the dilution formula.
c1v1 = c2v2
c = concn
v = volume
1N x 1mL = C2*20 mL
c2 = 1*1/20 = 0.2N after the first dilution. For the second we take 1 and dilute it to 50.
c1v1 = c2v2
0.2N x 1 mL = c2*50 mL
c2 = 0.2*1/50 = 0.004.
So there are 40 g in a liter of 1N solution. There will be 40 x 100/1000 mL = 4g in 100 mL.