Problem 1) A Beer\'s Law calibration was created for an ASA sample yielding the

Question

Problem 1) A Beer's Law calibration was created for an ASA sample yielding the following data.

Please attach the steps of how to solve these two problems.

Thank you

AO Chemistry 1034 STUDY GUIDE Page - 5 Practice Problems Problem - 1 A Beer's Law calibration curve was for created for an Aspirin (ASA) sample yielding the following data. Next a 10000 ml sample of ASA was prepared. A was 2.00 mL aliquot taken from this sample and diluted to 50.00 mLs. The for this sample absorbance was 0.129. Determine the strength of the aspirin tablet. MM(ASA) = 180.16 g/mole Beer's Law: Aspirin Analysis o 0 0 0 LA SA 0 mL ala 0.900 y = 666.68x +0.0073 o0 mls 0.800 0.700 R * = 0.9897 8 0.600 S 0.500 2 0.400 8 0.300 2 0.200 0.100 0.000 0.00E+00 2.00E-04 4.00E-04 6,00E-04_8.00E-04 1.00E-03120E-03 1,40E-03 [FelH20)4SA] (M) ( l. 0 s - AN LA oblem – 2 An unknown volatile liquid has a normal boiling point of 80.7 °C. The vapor pressure of this unknown liquid at 25.0 °C is 0.223 atm. Determine the heat of vaporization of the unknown liquid.

Explanation / Answer

1) The absorbance of the final sample was 0.129

You have an absorbance versus concentration plot which has a straight line fit as

y= 666.68x + 0.0073

The absorbance is in th y axis and you have a y axis data. Substitue in this equation and solve for x

0.129 = 666.68x + 0.0073

0.129 - 0.0073 = 666.68x

x = 1.83 x 10-4M

this is the concentration in the 50mL sample from which the absorbance weas measured. SO the aspitin content in this is

1.83 x 10-4 moles/L x 0.050 L= 9.127 x 10-6 moles

this is the amount of aspitin in the 2 mL that was added to make this solution.

9.127 x 10-6 moles is present in 2mL

100 mL of the solution has how many moles (9.127 x 10-6 moles x 100)/2 = 4.56 x 10-4 moles

MW of aspirin is 180.16g/mol so 4.56 x 10-4 moles is 4.56 x 10-4 moles x 180.16g/mol = 0.0822 g

or 82.2 mg

2) Normal boiling point is 80.7 oC,

means at 80.7oC the vapour pressure is 1 atm

Vapour pressure at 25oC is 0.223 atm

Using Clausius-Clapeyron Equation

ln (P1 / P2) = (H / R) (1/T2 - 1/T1)

ln(1/0.223) = (H / 8.314) (1/(273+25) - 1/(273+80.7))

1.50 = (H / 8.314)(1/(298) - 1/(353.7))

1.50 = (H / 8.314)5.28 x 10-4

(H / 8.314) = 1.5/5.28 x 10-4

(H / 8.314) = 2838.5

H = 23599 J

H = 23.6kJ/mol

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