Asked by Chem3A

How many milliliters of 0.503 M NaNO3 contain 2.528 g of NaNO3?
Answered by Damon
NaNO3 = 23 + 14 + 3(16) = 85 g/mol
2.528/85 = .03 mol *1000 mL/.503mol
= 59.1 mL