A chemist makes a solution of Mg(NO3)2 by dissolving 21.3 g Mg(NO3)2 in water to make 100.0 mL of solution. What is the concentration of NO3− ions in the solution? Assume that Mg(NO3)2 is the only solute in the solution. The molar mass of Mg(NO3)2 is 148.33 g/mol.

Answer : The concentration of is, 2.88 M

Explanation : Given,

Mass of = 21.3 g

Volume of solution = 100.0 mL

Molar mass of = 148.33 g/mole

First we have to calculate the concentration of

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

Now put all the given values in this formula, we get:

The concentration of is, 1.44 M

Now we have to calculate the concentration of

As, 1 mole of dissociates to give 1 mole of ion and 2 mole of ion.

So, 1.44 M of dissociates to give 1.44 M of ion and (2 × 1.44 M = 2.88 M) of ion.

Thus, the concentration of is, 2.88 M

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