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

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 

Answer 2

Answer:

$M=1.02M$

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:

$M_{NO_3^-}=\frac{n_{NO_3^-}}{V_{solution}}$

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

$n_{NO_3^-}=32.0gMg(NO_3)_2*\frac{1molMg(NO_3)_2}{148.3gMg(NO_3)_2}*\frac{2molNO_3^-}{1molMg(NO_3)_2}\\n_{NO_3^-}=0.432molNO_3^-$

Finally, the requested molarity turns out into:

$M_{NO_3^-}=\frac{0.432molNO_3^-}{0.425L}\\M=1.02M$

Best regards.