What is the concentration of nitrate ion in a 425 mL solution containing 32.0 g of Mg(NO3)2 (M = 148.3)?
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:
\(M=1.02M\)
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:
\(n_{NO_3^-}=32.0gMg(NO_3)_2*\frac{1molMg(NO_3)_2}{148.3gMg(NO_3)_2}*\frac{2molNO_3^-}{1molMg(NO_3)_2}\ _{NO_3^-}=0.432molNO_3^-\)
Finally, the requested molarity turns out into:
\(M_{NO_3^-}=\frac{0.432molNO_3^-}{0.425L}\\M=1.02M\)
Best regards.