Mass of solute: How many grams of barium hydroxide `[Ba(OH)_(2)]` are requirred to prepare `1.75 L` of a `0.0500 M` solution of barium hydroxide?
Strategy: Find the number of moles solute by multiplying the molartiy of solution with its volume and then multiply the number of moles of solute with its molar mass.

Explanation

Step 1: Calculate the number of moles of solute.
Molar of `Ba(OH)_(2)` = (Molarity of solution)
`xx` (Volume of solution)
`= (0.0500 mol L^(-1)) (1.75L)`
`= 0.0875 mol Ba(OH)_(2)`
Step 2: Calculate the mass of solute.
Mass of solute = (Moles of solute) `xx` (Molar mass of solute)
`= (0.087 mol)(171.3g mol^(-1))`
Alterntively, in one set up,
`?g Ba (OH)_(2) = 1.75 L soln xx (0.0500 mol Ba (OH)_(2))/(1L soln)`
`xx (171.3g Ba(OH)_(2))/(1 mol Ba(OH)_(2))`
`= 14.9g Ba (OH)_(2)`