Answer:
(B.) What is the molar solubility of barium chloride, BaCl2 in water?
Explanation:
Molar solubility is the number of moles of a substance that can dissolve in a liter of solution to the point of the solution's saturation. It can be calculated stoichiometrically from a substance's solubility product constant in mol/L.
Since all the [tex]BaCl_{2}[/tex] reacted all the [tex]Na_{2} SO_{4}[/tex] from the information, we can easily assume all the substances were consumed in the reaction, and hence account for their purity. Furthermore, [tex]BaSO_{4}[/tex] is insoluble in water, the most probable scientific query would be the molar solubility of the [tex]BaCl_{2}[/tex] used in the experiment.
Answer:
The best question is A Is the BaCl2(s) used in the experiment pure?Explanation:
Step 1: Data given
Mass of the BaCl2 sample = 10.0 grams
Volume of water = 50.0 mL
We add excess Na2SO4
A precipitate BaSO4 will be formed
Step 2: The balanced equation
BaCl2(aq) → Ba^2+(aq) + 2Cl-(aq)
Ba^2+(aq) + SO4^2-(aq) →BaSO4(s)
Step 3: Calculate moles BaCl2
Moles BaCl2 = 10.0 grams / 208.23 g/mol
Moles BaCl2 = 0.048 moles
Step 4: Calculate moles Ba^2+
For 1 mol BaCl2 we have 1 mol Ba^2+
For 0.048 moles BaCl2 we have 0.048 moles Ba^2+
Step 5: Calculate mass Ba^2+
Mass Ba^2+ = moles Ba^2+ * molar mass Ba^2+
Mass Ba^2+ = 0.048 moles * 137.33 g/mol
Mass Ba^2+ = 6.59 grams
After measuring the mass of barium in BaSO4 we can determine if the BaCl2 was pure or not.
If the mass = 6.59 grams the BaCl2 was pure
If the mass <6.59 grams the BaCl2 wasn't pure
The best question is A Is the BaCl2(s) used in the experiment pure?
