Question

Given solutions of `(i)` `Al_2(SO_4)_3`; `(ii)` `BaCl_2`; `(iii)` `Na_2SO_4(aq)`; `(iv)` `"equal volumes of (ii) and (iii)"`........which preparation should exert the MOST osmotic pressure?

Answer 1

We gots `(i)` `Al_2(SO_4)_3`; `(ii)` `BaCl_2`; `(iii)` `Na_2SO_4(aq)`; `(iv)` `"equal volumes of (ii) and (iii)"`........I suspect it is solution `(iv)`

Explanation:

And `(iv)` a solution obtained by mixing equal volumes of `(ii)` and `(iii)`.

Now the highest osmotic pressure will be expressed by the solution which CONTAINS LEAST number of ions..........

And `(i)` has got 5 ions in solution, `2xxAl^(3+)`, and `3xxSO_4^(2-)`.................

And `(ii)` has got 3 ions in solution, `2xxCl^(-)`, and `Ba^(2+)`.................

And `(iii)` has got 3 ions in solution, `2xxNa^(+)`, and `SO_4^(2-)`.................

But `(iv)` mixes `(ii)` and `(iii)`, and you know, or should know that `BaSO_4` is pretty insoluble stuff. (All sulfates are soluble except for `HgSO_4`, and `CaSO_4`, and `BaSO_4`.

And thus, the following reaction occurs........

`Ba^(2+) + SO_4^(2-)rarr BaSO_4(s)darr`

In this solution, there remain 2 equiv each of `Na^+` and `Cl^-`. As far as I can tell (and you have not set out the problem as clearly as I would like), because `BaSO_4` precipitates, this solution, `(iv)`, has the LEAST number of ions in solution, and thus the highest osmotic pressure.