Question

assuming complete dissociation how many chloride ions are present per ml in a solution obtained by mixing 100 ml of 0.5 m bacl2, 100 ml of 0.2 m kcl and 100ml water​

Answer 1

Answer:

In 100ml of the solution the number of moles of BaCl2 will be 100*0.5*10^-3 which will be equal to 0.05.

In BaCl2 the number of Cl ions present will be 2*0.05 = 0.1mol.

Number of moles of KCl in 100ml of solution.

100*0.2*10^-3 = 0.02.

Now, the total volume after mixing will be 100 + 100 +100 = 300ml.

Total moles of chlorine ions will be = 0.1 + 0.02 = 0.12.

So, the number of moles of Cl ions per ml voume will  be = 0.12/300.

So the number of Cl ions will be = 0.12 * 6.023*10^23/300 = 2.4 * 10^20.