Astudent in lab prepared a cereal sample solution for spectroscopic analysis as described in the lab manual, and found the concentration of fe3+ in the sample to be 2.50 x 10−5 m. if her sample solution was prepared by diluting 25 ml of the original solution to a total volume of 100 ml, what is the concentration of the original solution?

A student in the lab prepared a cereal sample solution for spectroscopic analysis as described in the lab manual, and found the concentration of Fe3+ in the sample to be 2.50 x 10−5 M. The concentration of the original solution is 1.00 × 10⁻⁴ M.

The concentration of a solution is the estimate of how much solute has been dissolved in the solvent.

The molar concentration of Fe3+ =  2.50 × 10⁻⁵ M

∴

In 1 liter, there are 2.50 × 10⁻⁵ M  of Fe3+

Now in 100 mL which is equivalent to 0.1 L, the concentration of Fe3+ will be:

= 2.5 × 10⁻⁶ M

Since the total volume of the solution is 100 mL, the 2.5  × 10⁻⁵ M is also present in 25 mL of the stock(original) solution.

∴

= 1.00 × 10⁻⁴ M concentration of the original solution

Therefore, we can conclude that the concentration of the original solution is 1.00 × 10⁻⁴ M.

Learn more about the concentration of a solution here:

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The concentration of the original solution is 1x10*-4 M

Explanation:

[Fe3+} = 2.50 x 10−5 M

It means that in 1 L, I have 2,50 x 10−5 moles of cation.

These are the rules of three involved.

1 L 2,50 x 10−5 moles

0,1 L (0,1 L . 2,50 x 10−5 m) / 1L = 2,50 x 10−6 moles

As the total volume of solution is 100 ml, the 2,50 x 10−6 moles are also in 25 ml of the stock solution.

So:

0,025 L 2,50 x 10−6 moles

1 L (1L . 2,50 x 10−6 m) / 0,025 L = 1x10*-4 M

your answer would be “dry”