Question 3
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.
Molarity is given by:
Molarity = moles of solute / Volume of solution in litre
(a) Molar mass of Co(NO3)2.6H2O
= 59 + 2 (14 + 3 × 16) + 6 × 18 = 291 g mol - 1
∴Moles of Co(NO3)2.6H2O = 30 / 291 mol
= 0.103 mol
Therefore, molarity = 0.103 mol / 4.3 L
= 0.023 M
(b) Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol
∴ Number of moles present in 30 mL of 0.5 M H2SO4 = (0.5 X 30 ) / 1000 mol
= 0.015 mol
Therefore, molarity = 0.015 mol / 0.5 L
= 0.03 M
