Final answer:
The volume of a 0.29M barium chlorate solution that contains 150 g of barium chlorate is 1.70 liters, calculated by first determining the number of moles in 150 g and then using the molarity equation to find the volume.
Step-by-step explanation:
To calculate the volume of a 0.29M barium chlorate solution that contains 150 g of barium chlorate (Ba(ClO₃)₂), we need to determine the number of moles of barium chlorate in 150 g.
The molar mass of barium chlorate is:
Ba = 137.33 g/mol
Cl = 2 × 35.45 g/mol ≈ 70.9 g/mol
O = 6 × 16.00 g/mol ≈ 96.00 g/mol
Total = 137.33 + 70.9 + 96.00 ≈ 304.23 g/mol
The number of moles of barium chlorate is calculated as follows:
Number of moles = Mass (g) ÷ Molar mass (g/mol)
= 150 g ÷ 304.23 g/mol
≈ 0.493 moles (three significant figures)
Next, we use the molarity equation: Molarity (M) = moles of solute ÷ volume of solution in liters (L) and rearrange it to solve for volume: Volume (L) = moles of solute ÷ Molarity (M).
So,
Volume (L) = 0.493 moles ÷ 0.29 M
≈ 1.70 L (three significant figures)
Therefore, the volume of the 0.29M barium chlorate solution that contains 150 g of barium chlorate is approximately 1.70 liters.