How many milliliters (ml) of 5 M MgCl2 must be added to 25 ml of a 0.5 M MgCl2 solution
to produce 1500 ml with a final MgCl2 concentration = 1.75 M? How much water?
Let's see how much MgCl2 we need in the final solution:
1500 mL of 1.75 M solution contains
1.5 * 1.75 = 2.625 moles of solute
25 mL of 0.5M solution contains
.025 * 0.5 = .0125 moles of solute
So, we need to add 2.6125 moles of solute.
That means we need 2.6125/5 = .5225 L of 5M solution (522.5 mL)
So, now we have .025+.5225 = .5475L of solution. We want 1.5L, so we have to add
1.500-.5475 = .9525L or 952.5 mL of water
Well, let's see if I can calculate that for you without clowning around too much. To determine the amount of 5 M MgCl2 you need to add, we can use the equation:
(C1)(V1) + (C2)(V2) = (Cf)(Vf)
Where C1 and V1 are the initial concentration and volume, C2 and V2 are the concentration and volume of the 5 M MgCl2, Cf is the final concentration, and Vf is the final volume.
Given that C1 = 0.5 M, V1 = 25 ml, Cf = 1.75 M, and Vf = 1500 ml, we can solve for V2, which represents the volume of the 5 M MgCl2 to be added.
(0.5 M)(25 ml) + (5 M)(V2) = (1.75 M)(1500 ml)
12.5 + (5 M)(V2) = 2625
5 M(V2) = 2612.5
V2 ≈ 522.5 ml
So, you would need to add approximately 522.5 ml of the 5 M MgCl2 solution to achieve a final concentration of 1.75 M.
As for the amount of water, to find that out we can subtract the combined volume of the solutions (25 ml + 522.5 ml) from the final volume (1500 ml).
Amount of water = 1500 ml - (25 ml + 522.5 ml)
Amount of water ≈ 952.5 ml
Therefore, you would need to add roughly 522.5 ml of the 5 M MgCl2 solution and approximately 952.5 ml of water.
To solve this problem, we need to use the equation:
(C1)(V1) + (C2)(V2) = (C3)(V3)
where C1 and V1 are the concentration and volume of the first solution, C2 and V2 are the concentration and volume of the second solution, and C3 and V3 are the concentration and volume of the final solution.
Let's plug in the given values:
C1 = 5 M (concentration of the 5 M MgCl2 solution)
V1 = unknown (volume of the 5 M MgCl2 solution)
C2 = 0.5 M (concentration of the 0.5 M MgCl2 solution)
V2 = 25 ml (volume of the 0.5 M MgCl2 solution)
C3 = 1.75 M (final concentration of the MgCl2 solution)
V3 = 1500 ml (final volume of the MgCl2 solution)
Using the equation, we can solve for V1:
(5 M)(V1) + (0.5 M)(25 ml) = (1.75 M)(1500 ml)
5V1 + 12.5 = 2625
5V1 = 2625 - 12.5
5V1 = 2612.5
V1 = 2612.5 / 5
V1 = 522.5 ml
Therefore, we need to add 522.5 ml of the 5 M MgCl2 solution.
To determine the amount of water, we subtract the volume of the 5 M MgCl2 solution from the final volume:
Amount of water = V3 - V1
Amount of water = 1500 ml - 522.5 ml
Amount of water = 977.5 ml
Therefore, we need to add 977.5 ml of water.
To find out how many milliliters of 5 M MgCl2 must be added and how much water is needed, we can use the concept of dilution.
Dilution is the process of reducing the concentration of a solute in a solution. The formula to calculate the new concentration (C1) after dilution is:
C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
Let's solve the problem step-by-step:
Step 1: Calculate the amount of MgCl2 from the initial solution:
Initial concentration (C1) = 0.5 M
Initial volume (V1) = 25 ml
Amount of MgCl2 from the initial solution = C1 * V1
= 0.5 M * 25 ml
= 12.5 mmol
Step 2: Calculate the amount of MgCl2 needed in the final solution:
Final concentration (C2) = 1.75 M
Final volume (V2) = 1500 ml
Amount of MgCl2 needed in the final solution = C2 * V2
= 1.75 M * 1500 ml
= 2625 mmol
Step 3: Find the additional amount of MgCl2 required:
Additional amount of MgCl2 needed = Amount of MgCl2 needed in the final solution - Amount of MgCl2 from the initial solution
= 2625 mmol - 12.5 mmol
= 2612.5 mmol
Step 4: Calculate the volume of the 5 M MgCl2 solution:
Concentration of 5 M MgCl2 solution = 5 M
Volume of 5 M MgCl2 solution (V3) = Additional amount of MgCl2 needed / Concentration of the 5 M MgCl2 solution
= 2612.5 mmol / 5 M
= 522.5 ml
Therefore, you need to add 522.5 ml of the 5 M MgCl2 solution to the initial solution.
Step 5: Calculate the volume of water required:
Volume of water = Final volume - Volume of 5 M MgCl2 solution
= 1500 ml - 522.5 ml
= 977.5 ml
So, to produce 1500 ml of a solution with a final MgCl2 concentration of 1.75 M, you need to add 522.5 ml of the 5 M MgCl2 solution and 977.5 ml of water.