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Thank you for visiting If you have 36 6 mL of concentrated sulfuric acid and it is diluted to make up 1 5 L of 0 25 M sulfuric. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!

If you have 36.6 mL of concentrated sulfuric acid and it is diluted to make up 1.5 L of 0.25 M sulfuric acid, what was the molarity of the concentrated sulfuric acid?

A. 0.6 M
B. 0.45 M
C. 6 M
D. 0.04 M

Answer :

Final answer:

The molarity of the concentrated sulfuric acid was calculated to be 10.21M using the formula for dilution M1V1 = M2V2. Thus, none of the options provided matches this result.

Explanation:

The subject of this question is the preparation of diluted solutions and the calculation of molarity. To find the molarity (M) of the concentrated sulfuric acid, you can use the formula M1V1 = M2V2, where M1 is the molarity of the concentrated solution, V1 is the volume of the concentrated solution, M2 is the molarity of the diluted solution, and V2 is the volume of the diluted solution.

Given that the volume of diluted solution (V2) is 1.5L or 1500mL, the molarity (M2) is 0.25M, and the volume of the concentrated solution (V1) is 36.6mL. We substitute these values into the formula:

M1 * 36.6mL = 0.25M * 1500mL

Rearranging for M1 we get,

M1 = (0.25M * 1500mL) / 36.6mL = 10.21M

So, none of the given options match the calculated molarity of the concentrated sulfuric acid.

Learn more about Molarity here:

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