Question

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 1

Answer : The concentration of $NO_3^-$ is, 2.88 M

Explanation : Given,

Mass of $Mg(NO_3)_2$ = 21.3 g

Volume of solution = 100.0 mL

Molar mass of $Mg(NO_3)_2$ = 148.33 g/mole

First we have to calculate the concentration of $Mg(NO_3)_2$

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

Formula used :

$\text{Molarity}=\frac{\text{Mass of }Mg(NO_3)_2\times 1000}{\text{Molar mass of }Mg(NO_3)_2\times \text{Volume of solution (in mL)}}$

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

$\text{Molarity}=\frac{21.3g\times 1000}{148.33g/mole\times 100.0mL}=1.44mole/L=1.44M$

The concentration of $Mg(NO_3)_2$ is, 1.44 M

Now we have to calculate the concentration of $NO_3^-$

As, 1 mole of $Mg(NO_3)_2$ dissociates to give 1 mole of $Mg^{2+}$ ion and 2 mole of $NO_3^-$ ion.

So, 1.44 M of $Mg(NO_3)_2$ dissociates to give 1.44 M of $Mg^{2+}$ ion and (2 × 1.44 M = 2.88 M) of $NO_3^-$ ion.

Thus, the concentration of $NO_3^-$ is, 2.88 M