Calculate the molarity of each of the ions in a solution when 3.0 litre of 4.0 M NaCl and 4.0 litre of 2.0 M $C a C l_{2}$ are mixed and diluted to 10 litre.

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Solution

moles of Na $^{+}$ ions =3x4 =12
Total volume = 10L
Molarity of Na $^{+}$ ions = $\frac{12}{10}$ =1.2M
Moles of Ca $^{2 +}$ ions = 2x4 =8
Therefore , molarity of Ca $^{2 +}$ ions =0.8M
Moles of Cl $^{-}$ ions =3x4 +2x4x2 =28
Molarity of Cl $^{-}$ ions = $\frac{28}{10}$ =2.8M

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