Calculate the molarity of each of the following solutions : (a) 30 g of `Co(NO_3)_2. 6H_2O` in 4.3 L of solution (b) 30 mL of 0.5 M `H_2SO_4` diluted to 500 mL.

Explanation

(a) Molar mass of` Co(NO_3)_2.6H_2 O = 58.7 + 2(14 + 48) + 6 xx 18 = 310.7 g "mol"^(-1)`
Number of moles of `Co(NO_3)_(2.6)H_2O = (30g)/(310 g "mol"^(-1) ) = 0.0966 `mol
Volume of solution = 4.3 L
Molarity of solution = 0.0966/4.3 = 0.022 M.
(b) 1000 mL of 0.5 M `H_2SO_4` contain `H_2SO_4` = 0.5 mole
` therefore `30 mL of 0.5 M `H_2SO_4` contain `H_2SO_4`= 0.5/1000 x 30 mole = 0.015 mole
Volume of solution = 500 mL = 0.500 L
Molarity of solution = Number of moles of solute/Volume of solution in L
`= (0.015)/(0.500) = 0.03 M`