Question

What volume in L of a 0.32 M Mg(NO3)2 solution contains 45 g of Mg(NO3)2?

Answer 1

Answer: 0.94L

Explanation:

As a solution of concentration of 0.32 M indicates that contains 0.32 moles of the compound in 1L of solution. So we need to calculate the grams of Mg(NO3)2 in 0.32 M:

$\frac{0.32 moles Mg(NO3)2}{L}$×$\frac{148.3 g Mg(NO3)2}{molMg(NO3)2}$$=\frac{47.5 g Mg(NO3)2}{L}$

This is telling you that you have 47.5 g de Mg(NO3)2 per liter of solution.

Now, using a rule of three you can know de volume that occupies a solution containing 45 g of Mg(NO3)2:

47.5 g Mg(NO3)2 → 1L solution

45 g Mg(NO3)2 → x

$x=\frac{45g * 1L}{47.5 g} = 0.94L$

Finally, we have 45g of Mg(NO3)2 in a volume of 0.94L of solution 0.32 M Mg(NO3)2.

Answer 2

First we need to find the moles of Mg(NO_3)_2,

$45g Mg(NO_3)_2\times\frac{1mol Mg(NO_3)_2}{148.32g}= 0.3034 mol$

Using the equation for molarity, $M=\frac{n}{V}$, where M is molarity n is number of moles and V is volume. We make V subject and solve

$M=\frac{n}{V}\\V=\frac{n}{M}= \frac{0.3034mol}{0.32mol/L}=0.94L$