At 30.0°C, the molar solubility of barium sulfate (BaSO₄) in water is 1.20×10⁻⁵ M. Calculate the solubility in grams per liter.

• A. 2.79×10⁻³ g/L
• B. 4.56×10⁻³ g/L
• C. 1.20×10⁻⁵ g/L
• D. 5.85×10⁻³ g/L

Answer: (A)

Verified step by step guidance

1. First, understand that molar solubility refers to the number of moles of solute that can dissolve in a liter of solution to reach saturation. Here, the molar solubility of BaSO₄ is given as 1.20×10⁻⁵ M.
2. To convert molar solubility to solubility in grams per liter, use the formula: solubility (g/L) = molar solubility (mol/L) × molar mass (g/mol).
3. Calculate the molar mass of barium sulfate (BaSO₄). The molar mass is the sum of the atomic masses of barium (Ba), sulfur (S), and oxygen (O). Use the periodic table to find these values: Ba = 137.33 g/mol, S = 32.07 g/mol, O = 16.00 g/mol. Since there are four oxygen atoms, multiply the atomic mass of oxygen by four.
4. Add the atomic masses together to find the molar mass of BaSO₄: Molar mass = 137.33 g/mol (Ba) + 32.07 g/mol (S) + 4 × 16.00 g/mol (O).
5. Finally, multiply the molar solubility by the molar mass to find the solubility in grams per liter: Solubility (g/L) = 1.20×10⁻⁵ mol/L × molar mass of BaSO₄.