Instrumental Analysis Midterm questions with
verified answers
A photometer with a linear response to radiation gave a reading of 625
mV with a blank in the light path and 149 mV when the blank was
replaced by an absorbing solution. Calculate
a. the percent transmittance and absorbance of the absorbing solution.
b. the expected percent transmittance if the concentration of absorber
is one half that of the original solution.
c. the percent transmittance to be expected if the light path through
the original solution is doubled. Ans✓✓✓ a. %T = (149/625) x 100 =
23.84%; A = -log(0.2384) = 0.622694
b. %T = 10^(-0.622684 ÷ 2) x 100 = 48.8262%
c.) %T = 10^(-0.622684 x 2) x 100 = 5.6835%
A portable photometer with a linear response to radiation registered
75.2 μA with a blank solution in the light path. Replacement of the
blank with an absorbing solution yielded a response of 23.7 μA
Calculate
a. the percent transmittance of the sample solution.
b. the absorbance of the sample solution.
c. the transmittance to be expected for a solution in which the
concentration of the absorber is one third that of the original sample
solution.
, d. the transmittance to be expected for a solution that has twice the
concentration of the sample solution. Ans✓✓✓ a. %T = (23.7 ÷ 75.2) x
100 = 31.516%
b. A = -log(0.31516) = 0.501
c. A = 1/3 x 0.501 = 0.167 → %T = 10^-0.167 x 100 = 68.1%
d. A = 2 x 0.501 = 1.002 → %T = 10^-1.002 x 100 = 9.95%
Calculate the frequency in hertz of
a. an X-ray beam at a wavelength of 2.65 Å.
b. an emission line of manganese at 403.1 nm.
c. the line at 694.3 nm produced by a ruby laser.
d. an infrared absorption peak at 9.6 Ans✓✓✓ For all of these, use the
formula ν = c/λ
a. ν = (3.00 x 10^8 m/s) ÷ (2.65 x 10^-9 m) = 1.13208 x 10^17 Hz
b. ν = (3.00 x 10^8 m/s) ÷ (403.1 x 10^-9 m) = 7.44232 x 10^14 Hz
c. ν = (3.00 x 10^8 m/s) ÷ (694.3 x 10^-9 m) = 4.3209 x 10^14 Hz
d. ν = (3.00 x 10 ^8 m/s) ÷ (9.6 x 10^-6 m) = 3.125 x 10^13 Hz
Describe the differences between the following pairs of terms, and list
any particular advantages possessed by one over the other:
a. phototubes and photomultiplier tubes
b. filters and monochromators as wavelength selectors. Ans✓✓✓ a. A
phototube is a simple vacuum tube that is sensitive to radiation. A
photomultiplier tube is much more complex. Unlike the phototube
verified answers
A photometer with a linear response to radiation gave a reading of 625
mV with a blank in the light path and 149 mV when the blank was
replaced by an absorbing solution. Calculate
a. the percent transmittance and absorbance of the absorbing solution.
b. the expected percent transmittance if the concentration of absorber
is one half that of the original solution.
c. the percent transmittance to be expected if the light path through
the original solution is doubled. Ans✓✓✓ a. %T = (149/625) x 100 =
23.84%; A = -log(0.2384) = 0.622694
b. %T = 10^(-0.622684 ÷ 2) x 100 = 48.8262%
c.) %T = 10^(-0.622684 x 2) x 100 = 5.6835%
A portable photometer with a linear response to radiation registered
75.2 μA with a blank solution in the light path. Replacement of the
blank with an absorbing solution yielded a response of 23.7 μA
Calculate
a. the percent transmittance of the sample solution.
b. the absorbance of the sample solution.
c. the transmittance to be expected for a solution in which the
concentration of the absorber is one third that of the original sample
solution.
, d. the transmittance to be expected for a solution that has twice the
concentration of the sample solution. Ans✓✓✓ a. %T = (23.7 ÷ 75.2) x
100 = 31.516%
b. A = -log(0.31516) = 0.501
c. A = 1/3 x 0.501 = 0.167 → %T = 10^-0.167 x 100 = 68.1%
d. A = 2 x 0.501 = 1.002 → %T = 10^-1.002 x 100 = 9.95%
Calculate the frequency in hertz of
a. an X-ray beam at a wavelength of 2.65 Å.
b. an emission line of manganese at 403.1 nm.
c. the line at 694.3 nm produced by a ruby laser.
d. an infrared absorption peak at 9.6 Ans✓✓✓ For all of these, use the
formula ν = c/λ
a. ν = (3.00 x 10^8 m/s) ÷ (2.65 x 10^-9 m) = 1.13208 x 10^17 Hz
b. ν = (3.00 x 10^8 m/s) ÷ (403.1 x 10^-9 m) = 7.44232 x 10^14 Hz
c. ν = (3.00 x 10^8 m/s) ÷ (694.3 x 10^-9 m) = 4.3209 x 10^14 Hz
d. ν = (3.00 x 10 ^8 m/s) ÷ (9.6 x 10^-6 m) = 3.125 x 10^13 Hz
Describe the differences between the following pairs of terms, and list
any particular advantages possessed by one over the other:
a. phototubes and photomultiplier tubes
b. filters and monochromators as wavelength selectors. Ans✓✓✓ a. A
phototube is a simple vacuum tube that is sensitive to radiation. A
photomultiplier tube is much more complex. Unlike the phototube
