Question

3.50 L of 0.200 M hydrochloric acid is prepared using 12.0 M HCl(aq)

stock solution to which water is added. What volume of the stock solution
is required?

a. 7.00 mL
b. 8.40 mL
c. 17.1 mL
d. 58.3 mL

Answer 1

To determine the volume of the stock solution required, we can use the formula:

Molarity1 x Volume1 = Molarity2 x Volume2

Where Molarity1 and Volume1 represent the initial concentration and volume, and Molarity2 and Volume2 represent the final concentration and volume.

Given:

Molarity1 = 12.0 M

Volume1 = ?

Molarity2 = 0.200 M

Volume2 = 3.50 L

Plugging in the values into the formula, we have:

12.0 M x Volume1 = 0.200 M x 3.50 L

Simplifying the equation, we can solve for Volume1:

Volume1 = (0.200 M x 3.50 L) / 12.0 M

Volume1 ≈ 0.0583 L

To convert this to milliliters, we multiply by 1000:

Volume1 ≈ 58.3 mL

Therefore, the volume of the stock solution required is approximately 58.3 mL.

The closest answer option is d. 58.3 mL.

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

Answer 2

Step-by-step explanation:

To determine the volume of the stock solution required to prepare 3.50 L of 0.200 M hydro chloric acid, we can use the formula:

M1V1 = M2V2

where:

M1 = concentration of the stock solution

V1 = volume of the stock solution

M2 = desired concentration of the diluted solution

V2 = desired volume of the diluted solution

Let's substitute the given values into the formula:

M1 = 12.0 M

V1 = ?

M2 = 0.200 M

V2 = 3.50 L

Now we can solve for V1:

12.0 M x V1 = 0.200 M x 3.50 L

V1 = (0.200 M x 3.50 L) / 12.0 M

V1 = 0.0583 L

To convert the volume from liters to milliliters, we multiply by 1000:

V1 = 0.0583 L x 1000 mL/L

V1 = 58.3 mL

Therefore, the volume of the stock solution required is 58.3 mL.

So, the correct answer is d. 58.3 mL.