Question

`100 ml` of `H_2SO_4`solution having molarity `1 M` and density `"1.5 g/ml"` is mixed with `400 ml` of water. Calculate final molarity of `H_2SO_4` solution, if final density is `"1.25 g/ml"`?

Answer 1

`M=0,227 (mol)/L`

Explanation:

volumes of different solutions are not additive, so 100 mL of sulfuric acid and 400 mL of water don't make 500 mL of solution but less.
in sp8ite of this you have 150 g of acid 1M + 400 g of water= 550 g of material and a final density of 1,25 g/mL then you have `V= M/d = (550g)/(1,25 g/(mL)) =440 mL`
At the beginning you have
`n= V/M = (0,1 L)/((1 mol)/L)= 0,1 mol`
of acid. In the final solution, you have the same amount then Molarity is `M=((mol)/V) = (0,1 mol)/(0,44L)= 0,227 (mol)/L`
pay attention: the utilizzed numbers are not real, but they are used only for this exercise