Question

A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of CaCl2 in this solution is M

Answer 1

You add 23.7 g of CaCl2 to 375 g of water. The total mass of the resulting solution is 398.7 g (CaCl2 + water).

Let's pretend we have 100 ml again. The density of the final solution is 1.05 g/ml. Therefore the mass of 100 ml = 100 x 1.05 = 105 g

You know that the ratio of

mass CaCl2 : mass total solution

is

23.7 : 398.7

You can use this ratio to work out the mass of CaCl2 in 105 g (100ml) of solution because the ratio is constant for both.

(mass CaCl2 / 105) = 23.7 / 398.7

therefore mass CaCl2 = (23.7 / 398.7) x 105

therefore the mass of CaCl2 in 105 g (100 ml) = 6.2415 g

So, you have 6.2415 g of CaCl2 in 100 mL of solution

moles of CaCl2 in 6.2415 g = mass / molecular weight

molecular weight CaCl2 = 40.08 + (2 x 35.45) = 110.98 g/mol

So moles = 0.0562 moles of CaCl2 in 6.2415 g,

thus 0.0562 moles in 100 ml of solution.

Molarity = moles / litres

= 0.0562mol / 0.1L

= 0.562 M CaCl2

Now, since CaCl2 -------> Ca^2+ + 2 Cl-

1 mole CaCl2 dissociates to give 2 moles Cl-. So the molarity of the Cl- is twice that of the CaCl2

=1.12 M