Normally and molarity: An aqueous solution contains `9.50g` of barium hydroxide in `200 mL` of solution. Calculate (i) the molartiy and (ii) the normally of the solution.
Strategy : Convert grams of `Ba(OH)_(2)` to moles of `Ba(OH)_(2)` and then calculate molarity. Becuase each formula unit of `Ba(OH)_(2)` furnishes two `Oh` ions, 1 mol `Ba (OH)_(2) = 2 eq Ba(OH)_(2)`. Thus, normally is twice of molarity for `Ba (OH)_(2)` solution.

Explanation

Step 1: Calculate the number of moles of solute.
Moles `Ba(OH)_(2) = ("Gramms" Ba(OH)_(2))/("Molar mass" Ba(OH)_(2))`
`= (9.50g)/(171.36g mol^(-1))`
`= 0.0554 mol Ba(OH)_(2)`
Step 2: Calcualte the molarity of solution.
Molarity `(M) = ("Number of moles")/("Number of literts")`
`= (0.0554 mol)/(2.00 L) = 0.277 M Ba (OH)_(2)`
Step 3: Calculate the normally of solution.
`N = M xx ("no. eq")/("mol")`
`= (0.277 ("mol")/(L)) ((2 eq)/("mol"))`
`= 0.0554 (eq)/(L) = 0.0554 N`