A 3.31 x 10^-4 M solution of a new drug (273 g.mol^-1) has an absorbance of 0.75

Question

A 3.31 x 10^-4 M solution of a new drug (273 g.mol^-1) has an absorbance of 0.750 in a 1.00-cm cell. A 0.3 122 g tablet containing the drug is dissolved in water to a final volume of 500.0 mL, and has an absorbance of 0.654 under the same conditions. What is the mass percent of the drug in the tablet? 87.4 56.9 18.9 12.6 In its protonated form, bromothymol blue has a molar absorptivity of 13800 M^-1.cm^-1 at 435 nm. At the same wavelength, the deprotonated form of bromothymol blue has a molar absorptivity of 3410 M^-1.cm^-1. If a solution contains 2.80 x M protonated bromothymol blue and 5.00 x 10^-5 M deprotonated bromothymol blue, what is the absorbance of the solution n a 1.00-cm sample cell at 435 nm? 0.228 0.557 0.785 1.34 How much mental effort did you expend on this question? h Low 1 2 3 4 5 High

Explanation / Answer

Q23answer :-

Formula to calculate the absorbance is as follows

A=e*b*c

e = Molar absorptivity constant

Lets calculate the absorbance of the protonated form

A= 13800 M-1cm-1*1 cm*2.80*10^-5 M

A = 0.386

Now lets calculate the absorbance of the deprotonated form

A=e*b*c

A= 3410 M-1.cm-1 *1 cm *5.00*10^-5 M

A=0.1705

Total absorbance = 0.386 +0.1705 = 0.557

So the answer is option 2 that is 0.557

Q 24 solution :-

3.31*10^-4 M solution has absorbance of the 0.750

Therefore lets calculate the concentration from the 0.654 absorbance

0.654 * 3.31*10^4 M / 0.750 = 2.886*10^-4 M

Now using the concentration and volume lets find the moles of the drug

Moles = molarity * volume in liter

Moles = 2.886*10^-4 mol per L * 0.500 L

= 1.443*10^-4 moles

Now lets calculate the mass of the drug in the tablet

Mass = moles * molar mass

Mass = 1.443*10^4 mol * 273 g per mol

Mass = 0.0394 g

Now lets calculate the percent of the drug from the tablet

% drug = (mass of drug / mass of tablet )*100%

=(0.0394 g / 0.3122 g)*100%

= 12.6 %

Therefore the percent of the drug in the tablet = 12.6 %

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