Question

8. How many grams of potassium nitrate, KNO3, are needed to prepare 250.0 mL of a 3.00 x 10-2 M KNO3 solution?

Answer 1

Molar mass: 101.1032 g/mol

$Moles = Morality × Volume$

$Moles =3.00×10^{-2}×\frac{250.0}{1000}=0.0075moles$

Mass $=0.0075×101.1032=0.758grams$