To what volume should you dilute 50.0 mL of a 4.50 M KI solution so that 28.5 mL of the diluted solution contains 2.75 g of KI?
Convert 2.75 g to moles KI.
??moles = grams/molar mass.
Then M KI is ?? moles/0.0285 L = ?M
You want to dilute the 4.50 M KI solution starting with 50 mL and you want the final solution to be ?M
4.50M x (50mL/x mL) = ?M
Solve for x in mL.
Check my thinking. Check the work.
387
To determine the volume to which you should dilute the 50.0 mL of a 4.50 M KI solution, you can use the equation for dilution:
M1V1 = M2V2
Where:
M1 = initial molarity of the solution
V1 = initial volume of the solution
M2 = final molarity of the solution
V2 = final volume of the solution
From the problem statement, we know:
M1 = 4.50 M
V1 = 50.0 mL
M2 = ?
V2 = 28.5 mL
First, we need to calculate the amount of KI in grams for the 28.5 mL solution. Since the molarity (M) is moles per liter (mol/L), we can use the formula:
Moles = mass / molar mass
The molar mass of KI is:
KI: 39.10 g/mol + 126.9 g/mol = 166.0 g/mol
Now, using the equation above, we can calculate the number of moles of KI in 2.75 g:
Moles = 2.75 g / 166.0 g/mol
Next, we can use the equation for dilution to find the final molarity (M2):
M1V1 = M2V2
(4.50 M)(50.0 mL) = M2(28.5 mL)
M2 = (4.50 M)(50.0 mL) / 28.5 mL
Finally, we can substitute the known values into the equation:
M2 = (4.50 M)(50.0 mL) / 28.5 mL
M2 = 7.89 M
Therefore, you need to dilute the 50.0 mL of a 4.50 M KI solution to a final volume of 28.5 mL to obtain a final concentration of 7.89 M.
To solve this question, we need to understand the concept of dilution and use some basic calculations.
Dilution is the process of reducing the concentration of a solute in a solution by adding more solvent. When performing a dilution, the number of moles of the solute remains constant, but the volume increases, resulting in a decrease in concentration.
Here are the steps to solve this problem:
1. Calculate the number of moles of KI in the 28.5 mL diluted solution using the given mass of KI.
Given:
Mass of KI = 2.75 g
The molar mass of KI (potassium iodide) is approximately 166 g/mol.
Number of moles of KI = Mass of KI / Molar mass of KI
= 2.75 g / 166 g/mol
≈ 0.0166 mol (rounded to four decimal places)
2. Use the stoichiometry of the balanced chemical equation involving KI, if provided, to determine the moles of KI in the original solution. However, since no such information is given, we can assume that the number of moles of KI is the same in both the original concentrated solution and the diluted solution.
Moles of KI in original solution = Moles of KI in diluted solution
= 0.0166 mol
3. Calculate the volume of the original concentrated solution needed to obtain the desired number of moles of KI. This can be done using the dilution equation:
Concentration1 × Volume1 = Concentration2 × Volume2
Where:
Concentration1 is the initial concentration of the solution (4.50 M),
Volume1 is the initial volume of the solution (50.0 mL),
Concentration2 is the final concentration of the solution (unknown),
Volume2 is the final volume of the solution (28.5 mL).
Rearranging the equation to solve for Volume2 (final volume):
Volume2 = (Concentration1 × Volume1) / Concentration2
Plugging in the values:
Volume2 = (4.50 M × 50.0 mL) / Concentration2
4. Finally, substitute the calculated values into the equation and solve for Concentration2 (final concentration).
Volume2 = (4.50 M × 50.0 mL) / Concentration2
28.5 mL = (4.50 M × 50.0 mL) / Concentration2
Rearrange the equation to solve for Concentration2:
Concentration2 = (4.50 M × 50.0 mL) / 28.5 mL
Concentration2 ≈ 7.89 M (rounded to two decimal places)
Therefore, to dilute the 50.0 mL of a 4.50 M KI solution, you should add enough solvent to reach a final volume of approximately 28.5 mL. This will result in a diluted solution with a concentration of approximately 7.89 M.