What mass of nano3 is needed to make 3.5 l of 2.2 m solution? (na = 23.0g, n=14.0g, o=16.0g)
explain in steps

= 654.5 g NaNO3

Explanation:

Molarity = Moles/ volume

Volume = 3.5 L

Molarity = 2.2 M

Therefore;

Moles = Molarity × volume

          = 3.5 L × 2.2 M

          = 7.7 moles

But 1 mole of NaNO3 is 85 g/mol

Mass of NaNO3 = 85 g/mole × 7.7 moles

                          = 654.5 g NaNO3

wher is the rest of the question