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Analytical Instrumentation Questions and Answers

Answer: A

Transmittance is the ratio of transmitted to incident intensity, a number between zero and one or expressed as a percentage. The fraction absorbed is one minus it for a non-scattering sample, the logarithm of the ratio is absorbance, and photon energy is set by the wavelength. Because a detector measures intensity directly, transmittance is what an instrument actually observes and absorbance is computed from it. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

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Answer: B

Absorbance is minus the base-ten logarithm of transmittance, so a transmittance of a tenth is an absorbance of one and a hundredth is two. One minus transmittance is the fraction absorbed, which is a different quantity that is not proportional to concentration. The logarithm is what makes absorbance additive and linear in concentration, and that linearity is the whole reason the quantity is defined at all. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

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3.

A sample transmits \( 10 \) per cent of the incident light. Its absorbance is

Answer: C

Taking the negative logarithm of the transmittance:

\[ A = -\log_{10}(0.10) = 1.0 \]

giving 1.0. The value 0.9 is the fraction absorbed, which is not the same quantity, and 0.1 is the transmittance itself. Because each unit of absorbance is a tenfold reduction in transmitted light, an absorbance of three means only a thousandth of the light gets through - which is where stray light begins to dominate.

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4.

A solution has an absorbance of \( 0.30 \). Its transmittance is about

Answer: D

Inverting the logarithmic relation:

\[ T = 10^{-0.30} \approx 0.50 \]

giving about 50 per cent. Thirty per cent would confuse the absorbance's numerical value with a percentage. That 0.3 absorbance is almost exactly half the light is worth remembering, since it is the same 3 dB relationship that appears throughout engineering, expressed in a different unit.

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Answer: A

Absorbance is linear in concentration, so a calibration curve is a straight line through the origin and a doubling of concentration doubles the reading. Transmittance is what the detector actually measures but falls exponentially with concentration. Absorbance depends on path length directly, and it has no upper bound. A quantity chosen because it linearises a relationship is worth the extra computation, and here the computation is a single logarithm. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

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Answer: B

Absorbance is the product of the molar absorptivity, the path length through the sample and the concentration, which is why doubling either the cell length or the concentration doubles the reading. Temperature and wavelength affect the absorptivity rather than appearing separately, and source intensity cancels in the ratio. That the law is a simple product of three terms is what makes it so easy to use and so easy to misuse when its assumptions fail. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

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Answer: C

Molar absorptivity is a property of the substance at a given wavelength, expressed per mole per litre per centimetre, and it can range from a few units to a hundred thousand. It is not a concentration, a length or a scattering measure. A large absorptivity means a sensitive determination, which is why an analytical wavelength is chosen at an absorption maximum where the value is both large and flat.

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8.

A solution of \( 10^{-5} \ \mathrm{M} \) in a \( 1 \ \mathrm{cm} \) cell has \( \varepsilon = 10^{4} \). Its absorbance is

Answer: D

Multiplying the three terms of the law:

\[ A = \varepsilon b c = 10^{4} \times 1 \times 10^{-5} = 0.1 \]

giving 0.1. Because the units of absorptivity are chosen to cancel those of path length and concentration, the product is dimensionless as an absorbance must be - which is a quick check that the right units have been used.

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9.

Using a \( 5 \ \mathrm{cm} \) cell instead raises that absorbance to

Answer: A

Absorbance is proportional to path length:

\[ A = 10^{4} \times 5 \times 10^{-5} = 0.5 \]

giving 0.5. A longer cell is the standard way to measure a dilute solution, bringing the reading into the range where photometric error is least. That one of the three terms is under the operator's control, and can be changed by a factor of ten, is a practical lever worth remembering when a sample is too dilute to read reliably.

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Answer: B

At high concentration solute molecules are close enough to influence one another's electronic states, so the absorptivity is no longer a constant and the calibration curve bends towards the concentration axis. Path length is fixed by the cell, and detector saturation occurs at high transmittance rather than high concentration. A law derived on the assumption that absorbers act independently must fail when they stop doing so, and dilution is the remedy. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

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