Calculate the molarity of each of the following solution (a) 30 g of Co(NO₃)₂.6H₂O in 4.3 L solution (b) 30 mL of 0.5 MH₂SO₄ diluted to 500 mL.

solution;

Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.

(a) Mol. mass of $\mathrm{Co}\left({\mathrm{NO}}_3\right). 6{\mathrm{H}}_2\mathrm{O}$

               $$\begin{gathered} =58.9+(14+3\times 16)2+6\left(18\right) \\ =58.9+(14+48)\times 2+108 \\ =58.9+124+108 = 290.9 \end{gathered}$$

Moles of $\mathrm{Co}(\mathrm{NO}{)}_3.6{\mathrm{H}}_2\mathrm{O}$
                                       $=\frac{30}{290.9}=0.103 \mathrm{mol}.$
Volume of solution = 4.3 L
Molarity, 
          $$\begin{gathered} \mathrm{M}=\frac{\mathrm{Moles} \mathrm{of} \mathrm{solute}}{\mathrm{Volume} \mathrm{of} \mathrm{solution} \mathrm{in} \mathrm{litre}} \\ = \frac{103}{4.3} = 0.024 \mathrm{M} \end{gathered}$$

(b) Number of moles present in 1000 ml of 0.5M H₂SO₄= 0.5 mol
therefore number of moles present in 30ml of 0.5M H₂SO₄=$\frac{0.5\times 30}{1000}$mol =0.015mol
therefore molarity =0.015/0.5L 

thus molarity is 0.03M