Yahoo Answers: Answers and Comments for Please help me with two chemistry questions i m so lost! What is the concentration of sucrose in the final solution? [Chemistry]
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From saraflem
enUS
Sat, 18 Jan 2020 03:23:02 +0000
3
Yahoo Answers: Answers and Comments for Please help me with two chemistry questions i m so lost! What is the concentration of sucrose in the final solution? [Chemistry]
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From Anonymous: It's very simple.
concentration = 2.5 M...
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Sat, 18 Jan 2020 03:31:56 +0000
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.

From Roger the Mole: (26.9 mL x 2.50 M) / (50.0 mL) x (13.5 mL / 45...
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Sat, 18 Jan 2020 05:52:51 +0000
(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

From Captain Matticus, LandPiratesInc: Just work the first one stepbystep.
V[1] * ...
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Sat, 18 Jan 2020 03:40:47 +0000
Just work the first one stepbystep.
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.

From Dr W: *** # 1 ***
let's let
.. A = stock solutio...
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Sat, 18 Jan 2020 15:44:19 +0000
*** # 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 BeerLambert 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