Question

What is the concentration of nitrate ion in a 425 mL solution containing 32.0 g of Mg(NO3)2 (M = 148.3)?

Answer 1

`M=1.02M`

Step-by-step explanation:

Hello,

In this case, we compute the concentration as the molarity of the nitrate ion as the ratio between the nitrate ion moles and the volume of the solution as shown below:

`M_(NO_3^-)=(n_(NO_3^-))/(V_(solution))`

Now, the nitrate ion moles are computed considering that in mole of magnesium nitrate, there are two moles of the nitrate ion:

`n_(NO_3^-)=32.0gMg(NO_3)_2*(1molMg(NO_3)_2)/(148.3gMg(NO_3)_2)*(2molNO_3^-)/(1molMg(NO_3)_2)\\n_(NO_3^-)=0.432molNO_3^-`

Finally, the requested molarity turns out into:

`M_(NO_3^-)=(0.432molNO_3^-)/(0.425L)\\M=1.02M`

Best regards.

Answer 2

There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. Calculations are as follows:

Molarity of NO3 ion = 32.0 g Mg(NO3)2 ( 1 mol / 148.3 g ) ( 2 mol NO3 / 1 mol Mg(NO3)2) / .425 L
MOlarity of NO3 ion = 1.02 M