Question

This information is for questions 42 and.43. A 5.00 g sample of soil is thought to contain naphthalene, a polycyclic aromatic hydrocarbon with a molar mass of i 128.1g mol and a molecular structure consisting of 2 fused benzene rings. Naphthalene has a UV absorbance peak at 220 nrm | (ε = 112,000 cm-,M-1). To find out how much naphthalene the : sample contains, an analytical chemist extracts the sample with i50.0 mL of hexane. The concentration of the napthalene in the 50.0 mL extract is then determined by UV spectrophotometry at 20 nm using an absorbance cell with a 1.00 em path length. . 42. When the first sample extract is placed in this cell, it transmits only 30.0% as much of the 220 nm radiation as when only hexane is in the cell. What is the absorbance (A) of the sample extract? (A) 0.153 (B) 0.523 (C) 0.297 (D) 70.3 For a second naphthalene extract measured with the same sample cell, the absorbance was 0.187 at 220 nm. What was the concentration of the naphthalene in this extract? 43. (A) 3.33×10-1 M (C) 1.67 x 10-6 M (B) 3.33 x 103 M (D) 6.67 x 10-7 M

Answer 1

Q42

Recall that Absorabnce an tranmittance can eb realted as:

A = 2-log(T%)

A = 2-log(30 )

A = 0.5228

A = 0.523; choose B

Q43

Beer’s Law

A = e*l*C

A = absorbance of sample

e is the molar absorptivity , the typical units are 1/M-cm

l size of cuvette, typically reported in cm

c is the molar concentration, in mol per ltier or M

A = e*l*C

0.187 = 112000*1*C

C = 0.187 /112000 = 0.00000166

C = 1.66*10^-6 M

choose C