Question

If a `0.3*mol` quantity of calcium chloride is dissolved in a `1*L` water solvent what are the concentrations of `Ca^(2+)` and `Cl^-`?

Answer 1

`"Molar concentration"="Moles of solute"/"Volume of solution"`

Explanation:

Now calcium chloride speciates in aqueous solution according to the following chemical equation.....

`CaCl_2(s) stackrel(H_2O)rarrCa^(2+) + 2Cl^-`

And thus if `0.3*mol` of calcium chloride are dissolved, we gets `0.6*mol` of CHLORIDE ions in solution, and `0.3*mol` of calcium ions....

Capisce?

Answer 2

0.6

Explanation:

`"Molar concentration"="Number of moles"/"Volume"`

In your example:
`"Number of moles"=0.3 " mol"`
`"Volume"=1000 " ml"`

To change the volume into Liters, we have to divide the volume by thousand (1L contains 1000 ml):
`"Volume"=(1000 " ml")/ (1000 " ml/L") =1" L"`

`rarr "Molar concentration"=0.3/1=0.3 " mol/L"`

Since we are finding the molar concentration of chloride ions:
`CaCl_2rarrCa^(2+)+2Cl^-`

Therefore the concentration of Chloride ions is double the concentration of Calcium chloride:
`"Molar concentration of Chloride"= 2*0.3=0.6" mol/L"`