Question

500 ml of 0.2 M AlCl3 is mixed with 500 ml of 0.1 M MgCl2 solution. Then molarity of Cl- in final solution is?

Answer 1

the molarity of Cl^-1 in aluminium chloride is 3 x 0.2 = 0.6 
the molarity of chloride ion in MgCl2 = 2 x 0.1 = 0.2 
 after mixing the volume changes =( M1V1 + M2V2) / V1+ V2 = 500x0.6 + 500 x 0.2 / 1000 = 0.4

Answer 2

500ml of 0.2M AlCl3 :

number of mole of AlCl3 solution = volume of solution in litre × concentration

= (500/1000)L × 0.2M

= 0.1mol

In one mole of AlCl3, contains 3 moles Cl- ions.

so, number of mole of Cl- ions in AlCl3 solution = 3 × 0.1 = 0.3 mol.

500ml of 0.1M MgCl2:

number of mole of MgCl2 solution = volume of solution in litre × concentration

= (500/1000)L × 0.1M

= 0.05mol

in one mole of MgCl2, contains 2 moles of Cl- ions.

so, number of mole of Cl- ions in MgCl2 solution = 2 × 0.05 = 0.1mol.

now total number of mole of Cl- ions when both the solutions are mixed = 0.3 + 0.1 = 0.4mol

and total volume of solution = 500ml + 500ml = 1000ml = 1L

hence, concentration of Cl- ions = number of mole of Cl- ions/volume of solution in L

= 0.4mol/1L

= 0.4M