Question

0.6573 g of potassium iodate is dissolved and diluted to a volume of 100.0 mL. Then 25.00 mL of this stock solution is pipetted into a 100.0 mL volumetric flask which is filled to the mark with water. What is the concentration of the final standard solution in mol/L? Include units in your answer, and round the final answer to the correct number of significant figures.

Answer 1

Answer 2

First, let's calculate the number of moles of potassium iodate in the stock solution:

moles KIO3 = mass / molar mass moles KIO3 = 0.6573 g / 214.00 g/mol moles KIO3 = 0.0030692 mol

Next, let's calculate the concentration of the stock solution in mol/L:

concentration = moles / volume concentration = 0.0030692 mol / 0.100 L concentration = 0.0307 mol/L

Finally, let's calculate the concentration of the final solution after dilution:

moles before dilution = moles after dilution concentration before dilution x volume before dilution = concentration after dilution x volume after dilution 0.0307 mol/L x 25.00 mL = concentration after dilution x 100.0 mL

Solving for concentration after dilution, we get:

concentration after dilution = (0.0307 mol/L x 25.00 mL) / 100.0 mL concentration after dilution = 0.00768 mol/L

Therefore, the concentration of the final standard solution is 0.00768 mol/L.