The mass of solid BaCl₂ needed to prepare the solution is 10.4 g
We'll begin by calculating the number of mole of BaCl₂ in the solution.
Volume = 250 mL = 250 / 1000 = 0.25 L
Molarity = 0.2 M
Mole of BaCl₂ =?
Mole = Molarity x Volume
Mole of BaCl₂ = 0.2 × 0.25
Mole of BaCl₂ = 0.05 mole
- Finally, we shall determine the mass of 0.05 mole of BaCl₂
Mole of BaCl₂ = 0.05 mole
Molar mass of BaCl₂ = 137 + (35.5×2) = 208 g/mol
Mass of BaCl₂ =?
Mass = mole × molar mass
Mass of BaCl₂ = 0.05 × 208
Mass of BaCl₂ = 10.4 g
Thus, the mass of BaCl₂ needed to prepare the solution is 10.4 g
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