Calculate the molarity of each of the following solutions:
Question:

Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2. 6H2O in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.

Solution:

Molarity is given by:

Molarity $=\frac{\text { Moles of solute }}{\text { Volume of solution in litre }}$

(a) Molar mass of Co (NO3)2.6H2O = 59 + 2 (14 + 3 × 16) + 6 × 18

= 291 g mol−1

$\therefore$ Moles of $\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}=\frac{30}{291} \mathrm{~mol}$

= 0.103 mol

Therefore, molarity $=\frac{0.103 \mathrm{~mol}}{4.3 \mathrm{~L}}$

= 0.023 M

(b) Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol

$\therefore$ Number of moles present in $30 \mathrm{~mL}$ of $0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}=\frac{0.5 \times 30}{1000} \mathrm{~mol}$

= 0.015 mol

Therefore, molarity $=\frac{0.015}{0.5 \mathrm{~L}} \mathrm{~mol}$

= 0.03 M