A chemist wants to prepare a stock solution of H2SO4 so that a sample of 20.00mL will produce a solution with a concentration of .50 M when added to 100.0mL of water.

A)what should the molarity of the stock solution be?
B)If the chemist wants to prepare 5.00L of the stock solution from a concentrated H2S)4, which is the 18.0 M, what volume of concentrated acid should be used?

(please explain how you get the answer..I'm so confused, help!!)

A. The easy way is to use the dilution formula. Also you MUST assume that the volumes are additive.

mL1 x M1 = mL2 x M2
0.5 x 120 = 20 x M2
M2 = 0.5 x 120/20 = 3M

B. Use the dilution formula again.
18M x mL1 = 5,000 mL x 3M.
Solve for mL1 which is volume of th 18 M stuff.

A) To find the molarity of the stock solution, we can use the dilution formula: M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

In this case, we know the final concentration (M2) is 0.50 M, the initial volume (V1) is 20.00 mL, and the final volume (V2) is 100.0 mL + 20.0 mL = 120.0 mL.

Rearranging the formula, we have: M1 = (M2V2) / V1

Plugging in the values, we get: M1 = (0.50 M)(120.0 mL) / 20.0 mL = 3.00 M

So, the molarity of the stock solution should be 3.00 M.

B) To find the volume of concentrated acid needed to prepare 5.00 L of the stock solution, we can again use the dilution formula.

In this case, the final volume (V2) is 5.00 L, the initial volume (V1) is the volume of concentrated acid we want to find, the initial molarity (M1) is 18.0 M, and the final molarity (M2) is 3.00 M (from part A).

Rearranging the formula, we have: V1 = (M2V2) / M1

Plugging in the values, we get: V1 = (3.00 M)(5.00 L) / 18.0 M = 0.833 L

Converting 0.833 L to milliliters, we have: 0.833 L * 1000 mL/L = 833 mL

So, the volume of concentrated acid that should be used is 833 mL.

And that's how you calculate it! But hey, don't be too acid-tight about it. Just remember to be careful with your dilutions, or the solution might turn a bit sour.

To find the molarity of the stock solution, we can use the formula:

M1V1 = M2V2

where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

A) By substituting the given values, we can solve for M1:

M1 × 20.00 mL = 0.50 M × (20.00 mL + 100.0 mL)

M1 × 20.00 mL = 0.50 M × 120.00 mL

M1 = (0.50 M × 120.00 mL) / 20.00 mL

M1 = 3.00 M

Therefore, the molarity of the stock solution should be 3.00 M.

B) To determine the volume of concentrated H2SO4 needed, we can use the formula:

M1V1 = M2V2

where M1 is the initial molarity (given as 18.0 M), V1 is the initial volume (unknown), M2 is the final molarity (3.00 M), and V2 is the final volume (5.00 L).

Substituting the given values, we have:

18.0 M × V1 = 3.00 M × 5.00 L

V1 = (3.00 M × 5.00 L) / 18.0 M

V1 ≈ 0.833 L

Therefore, the chemist should use approximately 0.833 L (or 833 mL) of concentrated H2SO4 to prepare 5.00 L of the stock solution.

To solve this problem, there are a couple of steps you need to follow. Let's break it down:

A) To find the molarity of the stock solution, you need to use the dilution formula. The formula is:

M1V1 = M2V2

where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

In this case, the initial volume (V1) is the volume of the sample (20.00 mL) and the final volume (V2) is the volume of the sample plus the volume of water (100.0 mL).

We are given that the final molarity (M2) is 0.50 M. Rearranging the formula, we have:

M1 = (M2 x V2) / V1

Substituting the given values, we get:

M1 = (0.50 M x 120.0 mL) / 20.00 mL

M1 = 3.00 M

So, the molarity of the stock solution should be 3.00 M.

B) To find the volume of concentrated acid needed to prepare 5.00 L of the stock solution, you need to use the dilution formula again. This time, the given values are:

- The initial molarity (M1) of the concentrated acid is 18.0 M.
- The initial volume (V1) is what we are trying to find.
- The final molarity (M2) is the molarity of the stock solution, which we found to be 3.00 M.
- The final volume (V2) is 5.00 L.

Rearranging the formula, we have:

V1 = (M2 x V2) / M1

Substituting the given values, we get:

V1 = (3.00 M x 5.00 L) / 18.0 M

V1 = 0.833 L

So, the volume of concentrated acid that should be used is 0.833 L (or 833 mL).

I hope this explanation helps! Let me know if you have any further questions.