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