Question

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

Answer 1

M=0.02

Step-by-step 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