Question

2 solutions of 1M CaCl2 and 2M AlCl3 were mixed in volume ratio of 1:2, calculate the concentration of Cl minus ion in the resulting mixture.

Answer 1

5M of chlorine will present

Answer 2

Answer:The concentration the $Cl^-$ ions is 4.66 mol/L.

Explanation:

$CaCl_2(aq)\rightarrow Ca^{2+}(aq)+2Cl^-(aq)$

1 M $CaCl_2$ and 2M $AlCl_3$ were mixed in volume ratio of 1:2.

Let volume of $CaCl_2$ be x Liters and volume of $AlCl_3$  be 2x Liters.

1 mole of $CaCl_2$ gives 2 mol of $Cl^-$ ions.

In 1 M of of $CaCl_2$ solution = 2 moles of $Cl^-$ ions

So, in x L of 1 M $CaCl_2$ solution = 2x moles of $Cl^-$ ions.

$AlCl_3(aq)\rightarrow Al^{3+}(aq)+3Cl^-(aq)$

1 mole of $AlCl_3$ gives 3 mol of $Cl^-$ ions.

In 2 M of $AlCl_3$ solution = 6 moles of $Cl^-$ ions

So, in 2x L of 2 M $AlCl_3$ solution= 2x × 6 moles= 12x moles of $Cl^-$ ions

'Total moles of $Cl^-$ after mixing of solutions: (12x +2x) moles

Volume of after mixing = 1x L + 2x L = 3x L

$[Cl^-]=\frac{(2x+12x) mol}{3x L}=4.66 mol/L$

The concentration the $Cl^-$ ions ios 4.66 mol/L.