What is the quantity, in moles, of NaI present in 175.0 mL of a 0.230 M NaI solution?

• A. 0.175 mol
• B. 0.230 mol
• C. 0.0403 mol
• D. 0.0138 mol

Answer: (C)

Verified step by step guidance

1. Identify the given information: the volume of the NaI solution is 175.0 mL and the molarity (M) of the solution is 0.230 M. Remember that molarity is defined as moles of solute per liter of solution.
2. Convert the volume from milliliters to liters because molarity uses liters as the volume unit. Use the conversion factor: \$1\, \text{L} = 1000\, \text{mL}\(. So, calculate the volume in liters as \)V = \frac{175.0}{1000}$ L.
3. Use the molarity formula to find the number of moles of NaI: \(M = \frac{\text{moles of solute}}{\text{liters of solution}}\). Rearranged to find moles, it becomes \(\text{moles of solute} = M \times V\).
4. Substitute the known values into the equation: multiply the molarity (0.230 M) by the volume in liters (from step 2) to get the moles of NaI.
5. Interpret the result as the quantity of NaI in moles present in the 175.0 mL solution. This value represents how many moles of NaI are dissolved in that volume.