# Please help me with two chemistry questions i m so lost! What is the concentration of sucrose in the final solution?

26.9 mL of 2.50 M stock solution of sucrose is diluted to 50.0 mL. A 13.5 mL sample of the resulting solution is then diluted to 45.0 mL. A 20.0 mL sample of this solution is then diluted to 75.0 mL. What is the concentration of sucrose in the final solution?

The concentration of manganese in steel can be determined spectrophotometrically by dissolving the steel in acid and oxidizing Mn to MnO4−. The resulting solution is purple and the absorbance at 525 nm can be monitored. A standard solution containing 0.228 mM of Mn has an absorbance of 0.343 in a 1.00 cm cell. The absorbance of an unknown solution of Mn is 0.446. What is the molar concentration of Mn in the unknown solution?

If you can only answer one or the other that is fine! Thank you in advance for your help!

### 4 Answers

- Dr WLv 73 months ago
*** # 1 ***

let's let

.. A = stock solution

.. B = first dilution

.. C = second dilution

.. D = third dilution

.. S = sucrose

next, we use dimensional analysis

.. 26.9mL A.... 2.50mmol S... 13.5mL B... .20.0mL C.. . 0.1076 mmol

----- ---- ---- x ----- ----- ----- x ----- ---- ---- x ---- ----- --- = ----- ----- ----- = 0.108M

.. 50.0mL B... ... 1mL A.. .... .. 45.0mL C.. .75.0mL D.. .. .. . 1mL D

*** # 2 ***

for this problem we use the Beer-Lambert law

.. A = ε *L*c

where

.. A = absorbance

.. ε = a constant for the solute and wavelength of light

.. L = path length (1.00cm

.. c = concentration

assuming that ε and L don't change

.. A1/c1 = A1/c2

.. 0.446 * (0.228mM / 0.343) = 0.296M

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- Roger the MoleLv 73 months ago
(26.9 mL x 2.50 M) / (50.0 mL) x (13.5 mL / 45.0 mL) x (20.0 mL / 75.0 mL) = 0.108 M

- - - - - - - - - - - - - - - - - -

(0.446) x (0.228 mM / 0.343) = 0.296 mM

- Dr WLv 73 months agoReport
excellent!

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- 3 months ago
Just work the first one step-by-step.

V[1] * C[1] + V[2] * C[2] = V[3] * C[3]

Volume 1 * Concentration 1 + Volume 2 * Concentration 2 = Volume 3 * Concentration 3

Usually, in problems like this one, the idea is that Volume 3 is equal to the sum of Volumes 1 and 2. Also, since we're diluting with just regular water, we're setting the concentration of solution in the water to 0. I'll show it in the first step and drop it afterwards. I'll use the letter "c" as a placeholder for final concentration for that particular part of the problem.

26.9 * 2.5 + (50 - 26.9) * 0 = 50 * c

26.9 * 2.5 / 50 = c

26.9 / 20 = c

269 / 200 = c

134.5 / 100 = c

1.345 = c

13.5 * 1.345 = 45 * c

13.5 * 1.345 / 45 = c

135 * 0.1345 / 45 = c

3 * 0.1345 = c

0.39 + 0.0135 = c

0.4035 = c

20 * 0.4035 = 75c

80 * 0.4035 = 300 * c

8 * 0.4035 = 30 * c

8 * 0.1345 = 10 * c

1.04 + 0.0360 = 10c

1.076 = 10c

0.1076 = c

The final molarity is 0.108, to 3 sf.

On the 2nd problem, we need to figure out which numbers are important:

0.228 mM = Concentration 1

0.343 = Absorbance 1

Concentration 2 is unknown

0.446 = Absorbance 2

If I'm reading this correctly, and I most likely am not reading it correctly, then if we related concentration and absorbance indirectly, then we would get:

C1 / A1 = C2 / A2

That is, the Concentration to Absorbance ratio would be the same

0.228 / 0.343 = C / 0.446

0.228 * 0.446 / 0.343 = C

C = 228 * 446 * 10^(-6) / (343 * 10^(-3))

C = 228 * 446 * 10^(3 - 6) / 343

C = 296.46647230320699708454810495627.... * 10^(-3)

C = 0.29646647....

To 3 sf

0.296 mM

Like I said, I'm probably way off on this, but that's my best guess, given the information provided. Maybe the wavelength and 1.0 cm cell mentions are important, but I'm guessing that they're not, at least not for this particular problem.

Thank you so much for taking the time to explain things and help me out. I appreciate it so much

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- Anonymous3 months ago
It's very simple.

concentration = 2.5 M * 26.9 mL/50 mL * 13.5 mL/45 mL * 20 mL/75 mL

this is essentially using the standard dilution equation C1 * V1 = C2 * V2.

"26.9 mL of 2.50 M stock solution of sucrose is diluted to 50.0 mL"

is the first dilution calculation. The final concentration is the concentration of sucrose in the 13.5 mL then taken.

Then you just keep repeating it which gives you the same answer as the single equation I showed.

that's not the simle way, that's the LONG way.

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