What concentration of Cl– results when 993 mL of 0.521 M LiCl is mixed with 883 mL of 0.309 M MgCl2?
PLEASE HELP!
I don't know how to set this up
Well, it looks like you've got a "clownundrum" on your hands! Don't worry, I'm here to help you "clownquer" this problem with a touch of humor!
To solve this problem, we need to find the concentration of Cl– ions in the final solution. Let's break it down step by step:
Step 1: calculate the moles of LiCl and MgCl2 using the given concentrations and volumes.
Moles of LiCl = volume of LiCl solution (in L) x concentration of LiCl (in mol/L)
Moles of MgCl2 = volume of MgCl2 solution (in L) x concentration of MgCl2 (in mol/L)
Step 2: calculate the total moles of Cl– ions by adding the moles of LiCl and MgCl2.
Total moles of Cl– ions = moles of LiCl + moles of MgCl2
Step 3: calculate the final volume of the solution by adding the volumes of LiCl and MgCl2.
Final volume of the solution = volume of LiCl solution + volume of MgCl2 solution
Step 4: calculate the final concentration of Cl– ions by dividing the total moles of Cl– ions by the final volume of the solution.
Final concentration of Cl– ions = total moles of Cl– ions / final volume of the solution
Now, go ahead and use this "clownvoluted" approach to solve the problem! I "clownfidently" believe in you!
To find the concentration of Cl- in the final solution, you need to use the concept of moles and the relationship between volume, concentration, and moles.
First, let's find the moles of Cl- in each solution.
For LiCl:
Moles of LiCl = Volume of LiCl x Concentration of LiCl
Moles of LiCl = 993 mL x 0.521 M
Convert mL to L by dividing by 1000:
Moles of LiCl = 0.993 L x 0.521 M
For MgCl2:
Moles of MgCl2 = Volume of MgCl2 x Concentration of MgCl2
Moles of MgCl2 = 883 mL x 0.309 M
Convert mL to L by dividing by 1000:
Moles of MgCl2 = 0.883 L x 0.309 M
Next, we need to calculate the total moles of Cl- ions by adding the moles of Cl- from LiCl and MgCl2:
Total moles of Cl- = Moles of LiCl + Moles of MgCl2
Now, let's calculate the concentration of Cl- in the final solution:
Concentration of Cl- = Total moles of Cl- / Total volume of the solution
Since the total volume of the solution is the sum of the volumes of LiCl and MgCl2, convert the volumes to liters:
Total volume of the solution = 0.993 L + 0.883 L
Finally, plug in the values to calculate the concentration of Cl-:
Concentration of Cl- = (Moles of LiCl + Moles of MgCl2) / (0.993 L + 0.883 L)
Simplify the expression and calculate the concentration of Cl-.
To solve this problem, we need to use the principles of molarity, which is a measure of the concentration of a solute in a solution. Molarity (M) is defined as moles of solute divided by the volume of the solution in liters.
First, let's determine the number of moles of Cl- in each solution using the formula:
moles = molarity x volume (in liters)
For LiCl solution:
moles of Cl- in LiCl = 0.521 M x 0.993 L = 0.516 mols
For MgCl2 solution:
moles of Cl- in MgCl2 = 0.309 M x 0.883 L = 0.273 mols
Next, let's calculate the total moles of Cl- by adding the moles of Cl- from both solutions:
Total moles of Cl- = moles of Cl- in LiCl + moles of Cl- in MgCl2
Total moles of Cl- = 0.516 mols + 0.273 mols = 0.789 moles
Lastly, we need to calculate the final concentration of Cl- by dividing the total number of moles by the final volume of the solution:
final concentration of Cl- = moles of Cl- / total volume (in liters)
Total volume = volume of LiCl solution + volume of MgCl2 solution
Total volume = 0.993 L + 0.883 L = 1.876 L
final concentration of Cl- = 0.789 moles / 1.876 L = 0.420 M
Therefore, the concentration of Cl- in the final solution is 0.420 M.
Technically this can't be set up BECAUSE volumes are not additive; however, I'm sure the author of the problem meant to say to assume volumes were additive.
mols LiCl = M x L = ?
mols Cl^- from LiCl = same ?
mols MgCl2 = M x L = ?
mols Cl^- from MgCl2 = 2x that number.
Total mols Cl^- = sum of Cl^- from above.
Then M = mols/L solution.