100 mL of LiNO3, 0.241 M is mixed with 240 mL of 0.618 M Ca(NO3)2. What is the final concentration of NO3- in the solution?
0.943
Just remember the definition of molarity. It is # mols/Liter of solution. That equation can be rearranged to give #mols = M x L.
# mols LiNO3 = 0.241 M x 0.100 L = ??
# mols NO3^- from LiNO3 is the same.
# mols Ca(NO3)2 = 0.618 M x 0.240 L = ??
# mols NO3^- from Ca(NO3)2 is twice that.
Total mols NO3^- = mols from LiNO3 + mols from Ca(NO3)2 = ??
Then molarity NO3^- = mols NO3^-/L of solution.
The volume, of course, is 0.100 L + 0.240 L = ??
0.508M
.508
Well, the best way to handle this situation is to get your chemistry jokes ready! So, let's clown around with some calculations.
First, let’s find out the moles of NO3- in both solutions.
For LiNO3:
moles of LiNO3 = volume (in L) × concentration (in M)
moles of LiNO3 = 0.100 L × 0.241 M
For Ca(NO3)2:
moles of Ca(NO3)2 = volume (in L) × concentration (in M)
moles of Ca(NO3)2 = 0.240 L × 0.618 M
Now, let's find the total moles of NO3- in the solution:
total moles of NO3- = moles of LiNO3 + moles of Ca(NO3)2
Using Avogadro's number and dividing by the total volume, you can then find the final concentration of NO3- in the solution.
But hey, you don't need a lab coat to crack a joke! Just remember, when chemists die, they barium. They just don't react. 😄
To find the final concentration of NO3- in the solution, we need to calculate the moles of NO3- in both LiNO3 and Ca(NO3)2, and then add them together.
First, let's calculate the moles of NO3- in LiNO3:
Moles of LiNO3 = Volume (in liters) × Molarity
Volume of LiNO3 = 100 mL = 100/1000 = 0.1 L
Molarity of LiNO3 = 0.241 M
Moles of LiNO3 = 0.1 L × 0.241 M = 0.0241 moles of LiNO3
Next, let's calculate the moles of NO3- in Ca(NO3)2:
Moles of Ca(NO3)2 = Volume (in liters) × Molarity
Volume of Ca(NO3)2 = 240 mL = 240/1000 = 0.24 L
Molarity of Ca(NO3)2 = 0.618 M
Moles of Ca(NO3)2 = 0.24 L × 0.618 M = 0.14832 moles of Ca(NO3)2
Since there are two moles of NO3- ions per mole of Ca(NO3)2, we multiply the moles of Ca(NO3)2 by 2 to get the moles of NO3-:
Moles of NO3- = 0.14832 moles × 2 = 0.29664 moles of NO3-
Now, let's add the moles of NO3- from both compounds to get the total moles of NO3- in the solution:
Total moles of NO3- = Moles of LiNO3 + Moles of NO3-
Total moles of NO3- = 0.0241 moles + 0.29664 moles = 0.32074 moles
Finally, we need to calculate the final concentration of NO3-, which is given by the ratio of moles to volume:
Final concentration of NO3- = Total moles of NO3- / Total volume of solution
Total volume of solution = Volume of LiNO3 + Volume of Ca(NO3)2
Total volume of solution = 0.1 L + 0.24 L = 0.34 L
Final concentration of NO3- = 0.32074 moles / 0.34 L ≈ 0.944 M
Therefore, the final concentration of NO3- in the solution is approximately 0.944 M.