How many moles of Ba(NO3)2 are present in 0.25 L of a 2.00 M Ba(NO3)2 solution?

• A. 0.125 mol
• B. 2.00 mol
• C. 0.25 mol
• D. 0.50 mol

Answer: (D)

Verified step by step guidance

1. Identify the given information: the volume of the solution is 0.25 L and the molarity (M) of Ba(NO3)2 is 2.00 M. Molarity is defined as moles of solute per liter of solution.
2. Recall the formula relating molarity, moles, and volume: \(\text{Molarity} (M) = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\).
3. Rearrange the formula to solve for moles of solute: \(\text{moles of solute} = \text{Molarity} \times \text{volume}\).
4. Substitute the known values into the equation: \(\text{moles of Ba(NO3)2} = 2.00\, M \times 0.25\, L\).
5. Calculate the product to find the number of moles of Ba(NO3)2 present in the solution.