Given that the molar mass of nano3 is 85.00 g/mol, what mass of nano3 is needed to make 4.50 l of a 1.50 m nano3 solution? use mc031-1.jpg.

Ans: Mass of NaNO3 = 574 g

Given:

Volume of NaNO3, V = 4.50 L

Molarity of NaNO3, M = 1.50 M i.e. 1.50 moles/L

Molar mass of NaNO3 = 85.00 g/mol

To determine:

The mass of NaNO3

Explanation:

Molarity of a solution can be expressed as the number of moles of the solute in a liter volume of solution

Molarity  of NaNO3 = moles of NaNO3/volume of solution in L

1.50 moles/L = moles/4.50 L

moles NaNO3 = 1.50 *4.50 = 6.75 moles

Now, moles of NaNO3 = mass of NaNO3/molar mass NaNO3

mass of NaNO3 = moles * molar mass =

                          = 6.75 moles * 85.00 g/mole

                          = 573.75 g = 574 g      

- aluminum and copper chloride.

- the chloride ion to separate the aluminum from the oxygen so that the aluminum can react with the copper ions (and the water molecules).

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