Question

The solubility of Ba3(PO4)2 is 0.519 mg/l. How many milligrams and moles of Ba3(PO4)2 dissolve in 200 ml of solution?

a) 103.8 mg and 1.63 x 10⁻4 moles
b) 51.9 mg and 8.15 x 10⁻5 moles
c) 207.6 mg and 3.26 x 10⁻4 moles
d) 155.7 mg and 2.44 x 10⁻4 moles

Answer 1

The solubility of Ba3(PO4)2 is 0.519 mg/L. To find out how many milligrams and moles of Ba3(PO4)2 dissolve in 200 mL of solution, we first need to calculate the number of moles dissolved. The correct answer is (b) 51.9 mg and 8.15 x 10^-5 moles.

Step-by-step explanation:

The solubility of Ba3(PO4)2 is given as 0.519 mg/L. To find out how many milligrams and moles of Ba3(PO4)2 dissolve in 200 mL of solution, we first need to calculate the number of moles dissolved.

Given that the solubility is 0.519 mg/L, we can convert it to g/L by dividing by 1000: 0.519 mg/L ÷ 1000 = 0.000519 g/L.

Converting g/L to moles per liter using the molar mass of Ba3(PO4)2 (601.93 g/mol) gives us: 0.000519 g/L ÷ 601.93 g/mol = 8.62 x 10^-7 mol/L.

Finally, to find the number of moles dissolved in 200 mL (0.2 L) of solution, we multiply the concentration by the volume: 8.62 x 10^-7 mol/L × 0.2 L = 1.724 x 10^-7 moles.

Therefore, the correct answer is (b) 51.9 mg and 8.15 x 10^-5 moles.