What is the molarity of a solution in which 0.45 grams of sodium nitrate (NaNO3) are dissolved in 265mL of the solution?

A. 29.8M

B. 87.2M

C. 0.02M

D. 0.52M

M=0.02

Explanation:

85 g NaNO3 - > 1 mol NaNO3

0.45 g NaNO3 - >x

x = (0.45 g NaNO3 * 1 mol NaNO3) / 85 g NaNO3 x = 5.3 x 10^-3 mol

265 mL = 0.265 L

M = mol solute / L solution

M = 5.3 x 10^-3 mol NaNO3/0.265 L M=0.02