3 years ago

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

Chemistry

1 answer:

andrezito [222]3 years ago

5 0

Answer:

= 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

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