If `5.6 ` of KOH is present in (a) 500 mL and (b) 1 litre of solution, calculate the molarity of each of these solutions.

Explanation

Mass of KOH `= 5.6 g`
No. of moles `= (5.6)/(56) =0.1` mol
(i) Volume of the solution `= 500 ml = 0.5 L`
(ii) Volume of the solution = 1L
Molarity `= ("Number of moles of solute")/("Volume of solution (in L)") = (0.1)/(0.5) = 0.2 M`
(iii) Volume of the solution = 1L
Molarity `= (0.1)/(1) = 0.1 M`