Beer-Lambert Law & Spectroscopy practice questions

Question 1

A student uses different regions of the electromagnetic spectrum to investigate a sample of gaseous $\text{HCl}$. Which of the following correctly identifies the molecular process associated with the absorption of microwave radiation?

Accepted Answer

Molecular rotation

Answer

Molecular vibration

Answer

Valence electron transitions

Answer

Ionization of core electrons

Question 2

A student performs a gravimetric analysis experiment to determine the mass percent of $\text{Cl}$ in an unknown soluble salt mixture. The student dissolves a $0.500 \text{ g}$ sample of the mixture in distilled water and adds an excess of $\text{AgNO}_3\text{(aq)}$ to precipitate $\text{AgCl(s)}$. The precipitate is collected by filtration and weighed. Which of the following experimental errors would cause the calculated mass percent of $\text{Cl}$ in the mixture to be higher than the actual mass percent?

Accepted Answer

Failing to completely dry the $\text{AgCl(s)}$ precipitate before measuring its final mass

Answer

Adding an insufficient volume of $\text{AgNO}_3\text{(aq)}$, leaving some $\text{Cl}^-\text{(aq)}$ unreacted in solution

Answer

Spilling a small portion of the solution containing the dissolved salt mixture before adding $\text{AgNO}_3\text{(aq)}$

Answer

Rinsing the $\text{AgCl(s)}$ precipitate with distilled water while it is on the filter paper prior to drying

Question 3

The orbital diagram below models the electron configuration of a neutral nitrogen atom.

Which of the following claims correctly identifies the electronic state of the atom and describes the energy change that occurs as the atom returns to its ground state?

Accepted Answer

It represents an excited state, and a photon is emitted as the electron moves from the $3s$ orbital to a $2p$ orbital.

Answer

It represents an excited state, and a photon is absorbed as the electron moves from the $3s$ orbital to a $2p$ orbital.

Answer

It represents the ground state, and a photon is emitted as the electron moves from the $3s$ orbital to a $2p$ orbital.

Answer

It represents the ground state, and a photon is absorbed as the electron moves from a $2p$ orbital to the $3s$ orbital.

Question 4

A student performs paper chromatography on a mixture of two dyes, Dye 1 and Dye 2, using cellulose paper as the stationary phase and ethanol, $\text{C}_2\text{H}_5\text{OH}$ (a polar solvent), as the mobile phase. The experimental results are shown in the table below.

| Dye | Solvent Front Distance (cm) | Dye Distance (cm) |
|---|---|---|
| Dye 1 | $10.0$ | $8.0$ |
| Dye 2 | $10.0$ | $3.0$ |

If the student repeats the experiment using hexane, $\text{C}_6\text{H}_{14}$ (a nonpolar solvent), as the mobile phase, which of the following predictions regarding Dye 1 is correct, and why?

Accepted Answer

Dye 1 will travel a shorter distance because it forms weaker intermolecular attractions with nonpolar hexane than with polar ethanol.

Answer

Dye 1 will travel a greater distance because nonpolar hexane moves faster up the paper, carrying all solutes farther.

Answer

Dye 1 will travel a greater distance because it forms stronger London dispersion forces with hexane than the dipole-dipole forces it formed with ethanol.

Answer

Dye 1 will travel the same distance because the retention factor ($R_f$) depends only on the stationary phase and is independent of the mobile phase.

Question 5

A student prepares five standard solutions of $\text{Co(NO}_3\text{)}_2\text{(aq)}$ and measures the absorbance of each at a wavelength of $510\text{ nm}$ using a cuvette with a path length of $1.0\text{ cm}$. The resulting calibration plot is shown below. The student then measures an unknown $\text{Co(NO}_3\text{)}_2\text{(aq)}$ sample under the same conditions and records an absorbance of $0.30$. Based on the calibration plot, what is the concentration of $\text{Co(NO}_3\text{)}_2\text{(aq)}$ in the unknown sample?

Accepted Answer

$0.060\text{ M}$

Answer

$0.015\text{ M}$

Answer

$0.030\text{ M}$

Answer

$0.15\text{ M}$

Question 6

A student prepares a solution of a tracer dye used to monitor water flow. The student obtains the following data at the wavelength of maximum absorbance.

| Dye concentration | Cuvette path length | Absorbance |
|---|---|---|
| $2.00 \times 10^{-5}\text{ M}$ | $2.00\text{ cm}$ | $0.800$ |

The Beer-Lambert relationship is $A=\varepsilon bc$, where $\varepsilon$ is the molar absorptivity. What is the molar absorptivity of the dye?

Accepted Answer

$2.00 \times 10^4\text{ M}^{-1}\text{cm}^{-1}$

Answer

$2.00 \times 10^3\text{ M}^{-1}\text{cm}^{-1}$

Answer

$4.00 \times 10^4\text{ M}^{-1}\text{cm}^{-1}$

Answer

$2.00 \times 10^6\text{ M}^{-1}\text{cm}^{-1}$

Question 7

A student sets up a paper chromatography experiment to separate pigments extracted from spinach leaves. Polar filter paper is used as the stationary phase, and a nonpolar organic solvent mixture is used as the mobile phase. As shown in the diagram, the student adds solvent to the beaker such that the liquid level is above the baseline where the spinach extract was spotted. Which of the following best predicts and explains the effect of this procedural error on the experiment?

Accepted Answer

The pigments will dissolve directly into the bulk solvent at the bottom of the beaker rather than traveling up the paper by capillary action, preventing separation.

Answer

The calculated $R_f$ values for all pigments will increase because the higher initial solvent level accelerates capillary action along the paper strip.

Answer

The calculated $R_f$ values for all pigments will decrease because the excess solvent dilutes the mobile phase and lowers its affinity for the pigments.

Answer

Nonpolar pigments will successfully travel up the paper while polar pigments remain at the baseline, resulting in incomplete separation.

Question 8

A student determines the concentration of an unknown solution of $\text{CuSO}_4\text{(aq)}$ using a spectrophotometer set to a wavelength of $635\text{ nm}$. First, the student measures the absorbance of a $2.0 \times 10^{-3}\text{ M}$ standard $\text{CuSO}_4\text{(aq)}$ solution in a cuvette with a path length of $0.50\text{ cm}$, obtaining an absorbance of $0.20$. Next, the student measures the absorbance of the unknown $\text{CuSO}_4\text{(aq)}$ solution in a standard cuvette with a path length of $1.0\text{ cm}$ and records an absorbance of $0.60$. Based on these data, what is the molar concentration of the unknown $\text{CuSO}_4\text{(aq)}$ solution?

Accepted Answer

$3.0 \times 10^{-3}\text{ M}$

Answer

$1.5 \times 10^{-3}\text{ M}$

Answer

$6.0 \times 10^{-3}\text{ M}$

Answer

$1.2 \times 10^{-2}\text{ M}$

Question 9

A student performs an experiment to determine the molar enthalpy of combustion, $\Delta H_{comb}$, of ethanol, $\text{C}_2\text{H}_5\text{OH}(l)$. A known mass of ethanol is burned in a spirit burner to heat a measured mass of water in an open metal calorimeter. The student assumes that all heat released by the combustion is absorbed by the water and uses $q = m c \Delta T$ to calculate the experimental molar enthalpy of combustion. However, a portion of the heat released during combustion is transferred to the surrounding air and metal calorimeter cup rather than to the water. Which of the following claims correctly predicts the effect of this heat loss on the calculated magnitude of $\Delta H_{comb}$ and provides the correct justification?

Accepted Answer

The calculated magnitude of $\Delta H_{comb}$ is lower than the actual value because the measured $\Delta T$ of the water is smaller than if all heat were transferred to the water.

Answer

The calculated magnitude of $\Delta H_{comb}$ is higher than the actual value because the heat lost to the surroundings increases the measured final temperature of the water.

Answer

The calculated magnitude of $\Delta H_{comb}$ is lower than the actual value because the heat loss causes the calculated mass of ethanol burned to be greater than the actual mass consumed.

Answer

The calculated magnitude of $\Delta H_{comb}$ is higher than the actual value because the unmeasured heat absorbed by the cup is added directly to the calculated heat absorbed by the water.

Question 10

A student prepares a calibration curve of absorbance versus concentration for $\text{CuSO}_4\text{(aq)}$ using a spectrophotometer set to an appropriate wavelength. While measuring the unknown sample, the student forgets to wipe fingerprints and water droplets off the outside of the cuvette. Which of the following best explains how this error affects the calculated concentration of the unknown solution?

Accepted Answer

The calculated concentration will be falsely higher because scattering of light by the surface residue reduces light reaching the detector, producing a higher absorbance value.

Answer

The calculated concentration will be falsely lower because less light reaches the detector, which is recorded as a lower absorbance value.

Answer

The calculated concentration will be falsely lower because water droplets on the cuvette dilute the solution inside, decreasing the concentration of solute particles.

Answer

The calculated concentration will be falsely higher because the residue increases the light path length, which increases the molar absorptivity of the solute.

Question 11

A student plots the standard Gibbs free energy change, $\Delta G^\circ$, as a function of temperature, $T$, for a chemical reaction, as shown in the graph below. Based on the graph, which of the following claims about the reaction is correct?

Accepted Answer

The reaction is endothermic ($\Delta H^\circ > 0$) and becomes thermodynamically favorable at temperatures above $300 \text{ K}$ because $T\Delta S^\circ$ exceeds $\Delta H^\circ$.

Answer

The reaction is endothermic ($\Delta H^\circ > 0$) and becomes thermodynamically favorable at temperatures above $300 \text{ K}$ because $\Delta S^\circ < 0$.

Answer

The reaction is exothermic ($\Delta H^\circ < 0$) and is thermodynamically favorable at all temperatures below $300 \text{ K}$ because $\Delta S^\circ > 0$.

Answer

The reaction is exothermic ($\Delta H^\circ < 0$) and becomes thermodynamically unfavorable at temperatures above $300 \text{ K}$ because $T\Delta S^\circ$ exceeds $\Delta H^\circ$.

Question 12

A student titrates a sample of a monoprotic acid, $\text{HA(aq)}$, with standardized $\text{NaOH(aq)}$. The reaction is represented by

$\text{HA(aq)}+\text{OH}^-\text{(aq)} \rightarrow \text{A}^-\text{(aq)}+\text{H}_2\text{O(l)}$

| Quantity | Value |
|---|---:|
| Volume of acid sample | $25.00\text{ mL}$ |
| Concentration of $\text{NaOH(aq)}$ | $0.1000\text{ M}$ |
| Initial buret reading | $0.30\text{ mL}$ |
| Final buret reading | $25.30\text{ mL}$ |

Before the initial reading is recorded, an air bubble is trapped in the buret tip. During the titration, the bubble is expelled and the tip fills completely with the titrant. The student uses the difference between the final and initial buret readings to calculate the molarity of the acid. Which of the following correctly describes how the air bubble affects the calculated molarity of $\text{HA(aq)}$?

Accepted Answer

The calculated molarity is too high because the recorded volume includes titrant that fills the buret tip but does not enter the flask.

Answer

The calculated molarity is too high because expelling the air bubble increases the concentration of the $\text{NaOH(aq)}$ that subsequently enters the flask.

Answer

The calculated molarity is too low because some of the recorded titrant volume fills the buret tip, causing the student to use too few moles of $\text{NaOH}$ in the calculation.

Answer

The calculated molarity is unchanged because subtracting the initial buret reading from the final buret reading eliminates the effect of the air bubble.

Question 13

A student plans to separate an equimolar liquid mixture of ethanol, $\text{C}_2\text{H}_5\text{OH}$, and water, $\text{H}_2\text{O}$, by fractional distillation at $1\text{ atm}$.

| Component | Normal boiling point |
|-----------|----------------------|
| $\text{C}_2\text{H}_5\text{OH}$ | $78^\circ\text{C}$ |
| $\text{H}_2\text{O}$ | $100^\circ\text{C}$ |

The apparatus includes an electrically heated distillation flask, a packed fractionating column, a thermometer, a water-cooled condenser, and a receiving flask. Which proposed setup, operating procedure, and prediction will best produce an ethanol-enriched initial distillate while correctly monitoring the vapor temperature?

Accepted Answer

Use the packed column, center the thermometer bulb at the entrance to the condenser, heat slowly, and collect the initial fraction separately; predict \(0.50

Answer

Use the packed column, submerge the thermometer bulb in the boiling mixture, heat slowly, and collect the initial fraction separately; predict \(0.50

Answer

Use the packed column, center the thermometer bulb at the entrance to the condenser, heat rapidly, and collect the initial fraction separately; predict \(0.50

Answer

Use the packed column, center the thermometer bulb at the entrance to the condenser, heat slowly, and collect the initial fraction separately; predict $x_{\text{ethanol}}=1.00$.

Question 14

A student studies the decomposition of a hypothetical reactant, $\text{A(aq)} \rightarrow \text{products}$, at constant temperature. The student measures the reactant concentration at several times and plots the data as shown.

Which of the following correctly identifies the order of the reaction with respect to $\text{A}$ and justifies the answer?

Accepted Answer

The reaction is second order because the integrated second-order rate law predicts a linear relationship between $1/[\text{A}]$ and elapsed time.

Answer

The reaction is zero order because any straight-line plot against elapsed time indicates that the reactant is consumed at a constant rate.

Answer

The reaction is first order because the integrated first-order rate law predicts a linear relationship between $1/[\text{A}]$ and elapsed time.

Answer

The reaction order cannot be determined because neither the initial rate nor the value of the rate constant is provided.

Question 15

A student evaluates ammonium nitrate, $\text{NH}_4\text{NO}_3$, for use in an instant cold pack by dissolving a known amount of the solid in water. The dissolution is endothermic. Initially, the water, solid, and room are at the same temperature. The student performs the experiment in a covered, thin-walled vessel without an insulating jacket and calculates the enthalpy of dissolution using

$$q_{\text{solution}}=m_{\text{solution}}c_{\text{solution}}(T_f-T_i)$$

and

$$\Delta H_{\text{soln}}=-\dfrac{q_{\text{solution}}}{n_{\text{NH}_4\text{NO}_3}}$$

Assume that heat exchange with the room is the only significant source of error. Compared with the accepted positive value of $\Delta H_{\text{soln}}$, how will the calculated value be affected, and why?

Accepted Answer

It will be too low, because thermal energy enters the colder solution from the room, reducing the magnitude of the measured temperature decrease.

Answer

It will be too high, because thermal energy entering from the room makes $T_f-T_i$ more negative.

Answer

It will be too high, because the calculation counts thermal energy entering from the room as energy absorbed by the dissolution.

Answer

It will be too low, because some thermal energy released by the dissolution is transferred from the solution to the room.

Question 16

A student uses a volumetric pipet to transfer a $25.00\text{ mL}$ sample of an unknown monoprotic acid, $\text{HA(aq)}$, into a flask. The student titrates the sample with standardized $0.1000\text{ M}$ $\text{NaOH(aq)}$, using phenolphthalein as the indicator. The neutralization reaction has a $1:1$ mole ratio.

Near the endpoint, the student adds too much titrant, producing a persistent deep-pink solution rather than the first persistent pale-pink color. The student records $28.64\text{ mL}$ as the volume of titrant delivered and uses this volume to calculate the initial concentration of the acid. All other measurements are made correctly.

Which statement correctly describes how the error affects the calculated concentration of $\text{HA(aq)}$?

Accepted Answer

The calculated concentration is higher than the actual concentration because the recorded $\text{NaOH(aq)}$ volume is too large, causing too many moles of $\text{HA(aq)}$ to be inferred from the $1:1$ mole ratio.

Answer

The calculated concentration is lower than the actual concentration because, at fixed $\text{NaOH(aq)}$ concentration, a larger delivered volume corresponds to fewer moles of $\text{OH}^-\text{(aq)}$.

Answer

The calculated concentration is equal to the actual concentration because any $\text{OH}^-\text{(aq)}$ added after equivalence remains unreacted and therefore cannot affect the titration calculation.

Answer

The calculated concentration is higher than the actual concentration because the deep-pink color indicates that $\text{HA(aq)}$ remains in excess at the endpoint.

Question 17

A student determines the concentration of a monoprotic acid, $\text{HA(aq)}$, by titrating a $25.00\text{ mL}$ sample with $0.1000\text{ M}$ $\text{NaOH(aq)}$. The endpoint is assumed to coincide with the equivalence point.

Before each trial, the student rinses the buret with deionized water, drains it, and then fills it directly with the $\text{NaOH(aq)}$ titrant. A small, reproducible volume of water remains in the buret and mixes with the titrant. The student calculates the acid concentration using the labeled $0.1000\text{ M}$ titrant concentration and the recorded buret volume.

How does rinsing the buret only with deionized water affect the calculated concentration of $\text{HA}$?

Accepted Answer

The calculated concentration is too high because the residual water dilutes the $\text{NaOH(aq)}$, so a larger buret volume is required to deliver the equivalence-point moles of $\text{OH}^{-}\text{(aq)}$, and that larger volume is multiplied by $0.1000\text{ M}$.

Answer

The calculated concentration is too low because the residual water lowers the concentration of the $\text{NaOH(aq)}$ delivered from the buret.

Answer

The calculated concentration is too high because the residual water dilutes the acid in the flask, so more titrant must be added to neutralize the acid.

Answer

The calculated concentration is unchanged because adding deionized water does not change the number of moles of $\text{HA}$ originally present in the flask.

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