Question

What mass of sodium iodide, NaI, should be used in order to prepare 500 mL of a .675 M solution of sodium iodide?

a.) .820 g
b.) 50.6 g
c.) 202 g
d.) 336 g

Answer 1

To prepare a 0.675 M solution of sodium iodide in 500 mL, 0.3375 mol of NaI is required. This corresponds to 50.6 g when using the molar mass of NaI (149.89 g/mol).

Step-by-step explanation:

To calculate the mass of sodium iodide (NaI) needed to prepare a 0.675 M solution, we must first determine the number of moles of NaI required and then convert that amount to grams. The formula for molarity (M) is:

M = moles of solute / volume of solution in liters

Rearranging the formula to solve for moles:

moles of solute = M × volume of solution in liters

For the given molarity and volume:

moles of NaI = 0.675 M × 0.500 L = 0.3375 mol

Next, we convert moles to grams using the molar mass of NaI (approximately 149.89 g/mol):

mass of NaI = moles of NaI × molar mass of NaI

mass of NaI = 0.3375 mol × 149.89 g/mol = 50.6 g

Therefore, 50.6 g of NaI is needed to prepare 500 mL of a 0.675 M solution, making option (b) the correct answer.