Machine-translated by Google. Questions are authored in English.
Top

Spectroscopic Techniques Questions and Answers

41.

Quartz cells rather than glass are required for ultraviolet spectroscopy because glass

Answer: A

Correct answer: absorbs ultraviolet radiation

Ordinary glass is opaque below about 350 nm, so it would absorb the very radiation being measured and no light would reach the detector across most of the ultraviolet range. Fused quartz transmits down to about 190 nm and is used instead. Glass is perfectly adequate in the visible region, and plastic cells serve there for routine work.

Please log in or register to report this question.

42.

Radiation of wavelength 5 \( \mu \)m corresponds to a wavenumber of

Answer: B

Correct answer: 2000 cm\( ^{-1} \)

Wavenumber is the reciprocal of the wavelength expressed in centimetres:

\[ \bar{\nu} = \frac{1}{5 \times 10^{-4}\ \mathrm{cm}} = 2000 \ \mathrm{cm}^{-1} \]

The conversion is worth committing to memory as \( \bar{\nu} = 10^4/\lambda_{\mu m} \). Wavenumber is used in infrared spectroscopy because it is directly proportional to energy, so the axis is linear in the quantity that matters.

Please log in or register to report this question.

43.

A transmittance of 10 per cent corresponds to an absorbance of

Answer: C

Correct answer: 1.0

Ten per cent transmittance means \( T = 0.1 \), so

\[ A = \log\frac{1}{0.1} = \log 10 = 1.0 \]

The scale is logarithmic, so each further factor of ten removed from the beam adds one to the absorbance: 1 per cent transmittance gives 2, and 0.1 per cent gives 3. Above about 2 the transmitted light is so small that stray light and detector noise dominate, which is why samples are diluted into the 0.2 to 1.0 range.

Please log in or register to report this question.

Answer: D

An absorbance of two means the incident intensity has been divided by a hundred. Readings much above two are unreliable because so little light reaches the detector that stray light and noise begin to dominate the measurement.

Please log in or register to report this question.

45.

A solution of molar absorptivity 5000 L/mol\( \cdot \)cm at 2 \( \times \) 10\( ^{-5} \) M in a 1 cm cell gives an absorbance of

Answer: A

Correct answer: 0.10

Applying the Beer-Lambert law,

\[ A = \varepsilon c l = 5000 \times 2 \times 10^{-5} \times 1 = 0.10 \]

The units cancel exactly: litres per mole per centimetre, times moles per litre, times centimetres. A molar absorptivity of 5000 is moderately strong, and it means a solution of only twenty micromolar is measurable in an ordinary cell.

Please log in or register to report this question.

46.

If that solution's absorbance is 0.10 in a 1 cm cell, in a 4 cm cell it would be

Answer: B

Correct answer: 0.40

Absorbance is proportional to path length, so quadrupling the cell length quadruples it:

\[ A = 0.10 \times 4 = 0.40 \]

Long-path cells are used exactly this way to bring a weakly absorbing sample into the range where the measurement is most precise, and gas cells with multiple reflections achieve path lengths of many metres by the same principle.

Please log in or register to report this question.

47.

A solution giving an absorbance of 0.60 is diluted to one third of its concentration. Its absorbance becomes

Answer: C

Correct answer: 0.20

Absorbance is proportional to concentration, so diluting to one third divides it by three:

\[ \frac{0.60}{3} = 0.20 \]

This linearity is the whole basis of quantitative absorption spectrometry, and it is why absorbance rather than transmittance is the quantity plotted - transmittance would have gone from 25 to 63 per cent, which is not proportional to anything.

Please log in or register to report this question.

48.

Benzene's infrared spectrum shows far fewer bands than the 30 its atom count predicts because many modes are

Answer: D

Correct answer: degenerate or infrared inactive

Benzene has \( 3 \times 12 - 6 = 30 \) modes, but its high symmetry reduces the number of distinct bands enormously. Many modes come in degenerate pairs of identical frequency and so give one band between them, and many others produce no change in dipole moment and are infrared inactive altogether - though several of those appear in the Raman spectrum. Symmetry, not detector sensitivity, is the reason.

Please log in or register to report this question.

49.

Carbon dioxide, being linear with 3 atoms, has how many fundamental vibrational modes?

Answer: A

Correct answer: 4

Carbon dioxide is linear, so the formula is

\[ 3N - 5 = 3(3) - 5 = 4 \]

The four are the symmetric stretch, the asymmetric stretch, and two degenerate bends at right angles to each other. Only three frequencies appear, since the bends coincide, and only two bands are seen in the infrared, since the symmetric stretch produces no dipole change.

Please log in or register to report this question.

50.

A proton signal is split into 5 lines. The number of equivalent neighbouring protons is

Answer: B

Correct answer: 4

The multiplicity is \( n + 1 \), so five lines mean

\[ n = 5 - 1 = 4 \]

equivalent neighbouring protons, giving a quintet with intensities 1:4:6:4:1. The middle methylene of a propyl chain flanked by a methyl and another methylene shows exactly this pattern.

Please log in or register to report this question.

Loading…
Tags: Spectroscopic Techniques Questions and Answers || Spectroscopic Techniques MCQ Questions and Answers || Spectroscopic Techniques GK Questions and Answers || Spectroscopic Techniques GK MCQ Questions || Spectroscopic Techniques Multiple Choice Questions and Answers || Spectroscopic Techniques GK || GK on Spectroscopic Techniques || Engineering Chemistry Questions and Answers || Engineering Chemistry MCQ Questions and Answers || Engineering Chemistry GK Questions and Answers || GK on Engineering Chemistry