If one assumes the volumes are additive, what is the concentration, in mol/L, of NO3− in a solution obtained by mixing 263 mL of 0.167 M KNO3, 311 mL of 0.404 M Mg(NO3)2, and 713 mL of H2O?
In this problem, you need to find two things: the moles of solute (NO₃⁻) and the volume of the total solution.
First, find the moles of NO₃⁻ we have in each solution.
n₁ = (0.263 L)(0.167 M KNO₃) = 0.043921 mol KNO₃
Now, since there is one mole of NO₃⁻ per one mole of KNO₃, we can say easily say that the number of moles of NO₃⁻ is also 0.043921 mol.
n₂ = (0.311 L)(0.404 M Mg(NO₃)₂) = 0.125644 mol Mg(NO₃)₂
IMPORTANT: since there are TWO moles per one mole of Mg(NO₃)₂, we have to say that the number of moles of NO₃⁻ is DOUBLED the number of moles in Mg(NO₃)₂. Thus, it is 0.251288 mol NO₃⁻
Now we need to find the TOTAL number of moles of NO₃⁻.
n = 0.043921 mol + 0.251288 mol = 0.295209 mol NO₃⁻
Next, find the total volume. (pretty easy...)
V = 0.263 L + 0.311 L + 0.713 L = 1.287 L
Finally, use all the info we calculated to find the concentration of NO₃⁻ in the resulting solution.
M = (0.295209 mol) / (1.287 L) = 0.229 M NO₃⁻
Note: Notice how the final answer has only 3 sig figs. You do not round until the end. Remember that :)
Thus, the NO₃⁻ concentration of the resulting solution is 0.229 M NO₃⁻
mols NO3^- in KNO3 = M x L = ?
mols NO3^- in Mg(NO3)2 = 2*M x L = ?
M NO3^- = total mols/total L solution
Cheers, thanks man
To find the concentration of NO3− in the resulting solution, we need to consider the amounts of NO3− from the KNO3 and Mg(NO3)2 solutions and the total volume of the resulting solution.
First, we need to find the total amount of NO3− from each component of the mixture.
For the KNO3 solution:
Amount of NO3− from KNO3 = Volume of KNO3 solution × Concentration of KNO3
Amount of NO3− from KNO3 = 263 mL × 0.167 M = 43.921 mol
For the Mg(NO3)2 solution:
Amount of NO3− from Mg(NO3)2 = Volume of Mg(NO3)2 solution × Concentration of Mg(NO3)2 × (Number of NO3− ions from one formula unit of Mg(NO3)2)
Amount of NO3− from Mg(NO3)2 = 311 mL × 0.404 M × 2 = 251.192 mol
Next, we need to calculate the total volume of the resulting solution.
Total volume of resulting solution = Volume of KNO3 solution + Volume of Mg(NO3)2 solution + Volume of water
Total volume of resulting solution = 263 mL + 311 mL + 713 mL = 1287 mL
To convert the total volume to liters:
1287 mL = 1287 mL × (1 L / 1000 mL) = 1.287 L
Finally, we can calculate the concentration of NO3− in the resulting solution by dividing the total amount of NO3− by the total volume of the solution.
Concentration of NO3− in the resulting solution = (Amount of NO3− from KNO3 + Amount of NO3− from Mg(NO3)2) / Total volume of resulting solution
Concentration of NO3− in the resulting solution = (43.921 mol + 251.192 mol) / 1.287 L = 258.113 mol/L
Therefore, the concentration of NO3− in the resulting solution is approximately 258.113 mol/L.