One table of brand Y aspirin was dissolved in 10 mL of 1.0 M NaOH, boiled, and t

Question

One table of brand Y aspirin was dissolved in 10 mL of 1.0 M NaOH, boiled, and then diluted to a final volume of 100 mL with deionized water, A 2.5 mL aliquot of this concentrated aspirin solution was diluted to a final volume of 100 mL with FeCl_3, A calibration plot of absorbance vs concentration was obtained, with the slope of the best-fit straight line as 1370 m^-1. The diluted aspirin solution has an absorbance of 0.932. What is concentration on the dilute aspirin? What is the concentration ot the aspirin in the dilute aspirin solution? What is the concentration of the original aspirin solution? How many mg of aspirin are present in one tablet of Brand Y aspirin?

Explanation / Answer

a)

concentration of aspirin in dilute solution

Apply Beer Law

A = m*C

where m = 1370 M is given

an dA = 0.932 for our specific value

so

0.932 = 1370*C

C = 0.932/1370 = 0.000680 M

B)

then, in original concentration

Recall that

M1*V1 = M2*V2

Voriginal = 10 mL of NaOH

V2 = 100 mL

V3= 2.5 mL

V4 = 100 mL

So..

Find Concentratino in V3. 2.5 mL aliquot

M3*V3 = M4*V4

M3 = M4*V4/V3 = 0.000680*100/2.5 = 0.0272 M

now... this was from a set of 100 mL

so M2 = M3 since same solution

so

M1*V1 = M2*V2

M1*10 = 0.0272*100

M1 = 0.0272*100/10 = 0.272 M

c)

find mass of aspirin

so

M = mol/V

mol = MV = 0.272*(10*10 ^ -3) = 0.00272 moles of aspirin

mass = mol*MW = (0.00272*180) = 0.4896 g

mass in mg = 489.6 mg of aspirin

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