How much nano3 is needed to prepare 225 ml of a 1.55 m solution of nano3?

= 29.64 g NaNO3

Explanation:

Molarity is given by the formula;

Molarity = Moles/Volume in liters

Therefore;

Number of moles = Molarity * Volume in liters

= 1.55 M * 0.225 L

= 0.34875 moles NaNO3

Thus; 0.34875 moles of NaNO3 is needed equivalent to;

= 0.34875 moles * 84.99 g/mol

= 29.64 g