Portage Learning: CHEM 121 Lab Exam 2
Advanced Prep: Master Density, Specific
Heat, and Physical/Chemical Properties
Practice Questions & Detailed Explanations
Subject / Subtopic: General Chemistry I Lab (CHEM 121) / Density, Specific
Heat, and Physical vs. Chemical Properties
Question 1: A student is tasked with determining the density of an irregularly shaped, highly
porous ceramic material that reactively decomposes in water but remains completely inert in
cyclohexane ($\text{density} = 0.778\text{ g/mL}$). The dry mass of the ceramic sample is
$14.250\text{ g}$. To prevent the liquid from filling the internal pores during the displacement
measurement, the student coats the sample uniformly with a thin, water-insoluble polymer wax
($\text{density} = 0.920\text{ g/mL}$). The total mass of the coated ceramic sample is
$15.170\text{ g}$. When the coated sample is fully submerged in a graduated cylinder
containing cyclohexane, the liquid volume rises from $25.00\text{ mL}$ to $39.50\text{ mL}$.
Calculate the true density of the porous ceramic material itself.
A) $1.11\text{ g/mL}$
B) $1.06\text{ g/mL}$
C) $1.35\text{ g/mL}$
D) $0.982\text{ g/mL}$
Correct Answer: B) 1.06 g/mL
Explanation: To find the true density of the ceramic material, we must isolate its volume from the
volume of the polymer wax coating. First, determine the mass of the wax: $15.170\text{ g} -
14.250\text{ g} = 0.920\text{ g}$. Using the density of the wax, calculate its volume: $0.920\text{
g} / 0.920\text{ g/mL} = 1.000\text{ mL}$. Next, find the total displacement volume of the coated
sample: $39.50\text{ mL} - 25.00\text{ mL} = 14.50\text{ mL}$. Subtract the volume of the wax
to find the true volume of the ceramic material: $14.50\text{ mL} - 1.000\text{ mL} = 13.50\text{
mL}$. Finally, calculate the density of the ceramic: $14.250\text{ g} / 13.50\text{ mL} =
1.0556\text{ g/mL}$, which rounds to $1.06\text{ g/mL}$ based on significant figures. Option A
ignores the wax volume deduction. Option C miscalculates the mass-to-volume ratio, and Option
D fails to separate the mass and volume components accurately.
Question 2: In a calorimetry experiment designed to determine the specific heat capacity of an
unknown metal alloy, a $55.00\text{ g}$ sample of the alloy is heated to $99.50^\circ\text{C}$
and quickly transferred to a coffee-cup calorimeter containing $100.00\text{ g}$ of deionized
, water at $22.00^\circ\text{C}$. The highest equilibrium temperature reached by the system is
recorded as $26.80^\circ\text{C}$. If the calorimeter constant (heat capacity of the calorimeter
itself) is $12.5\text{ J/}^\circ\text{C}$, and the specific heat of water is $4.184\text{
J/g}^\circ\text{C}$, what is the specific heat capacity of the metal alloy?
A) $0.518\text{ J/g}^\circ\text{C}$
B) $0.473\text{ J/g}^\circ\text{C}$
C) $0.442\text{ J/g}^\circ\text{C}$
D) $0.534\text{ J/g}^\circ\text{C}$
Correct Answer: A) 0.518 J/g°C
Explanation: The heat lost by the metal alloy must equal the heat gained by both the water and
the calorimeter: $-q_{\text{metal}} = q_{\text{water}} + q_{\text{calorimeter}}$. Calculate heat
gained by water: $q_{\text{water}} = m \cdot c \cdot \Delta T = 100.00\text{ g} \cdot 4.184\text{
J/g}^\circ\text{C} \cdot (26.80^\circ\text{C} - 22.00^\circ\text{C}) = 100.00 \cdot 4.184 \cdot
4.80 = 2008.32\text{ J}$. Calculate heat gained by the calorimeter: $q_{\text{calorimeter}} =
C_{\text{cal}} \cdot \Delta T = 12.5\text{ J/}^\circ\text{C} \cdot 4.80^\circ\text{C} = 60.0\text{
J}$. Total heat gained = $2008.32\text{ J} + 60.0\text{ J} = 2068.32\text{ J}$. Therefore,
$q_{\text{metal}} = -2068.32\text{ J}$. Set up the equation for the metal: $-2068.32\text{ J} =
55.00\text{ g} \cdot c_{\text{metal}} \cdot (26.80^\circ\text{C} - 99.50^\circ\text{C})$. Thus, $-
2068.32 = 55.00 \cdot c_{\text{metal}} \cdot (-72.70)$, yielding $c_{\text{metal}} = 2068.32 /
3998.5 = 0.51727\text{ J/g}^\circ\text{C}$, or $0.518\text{ J/g}^\circ\text{C}$. Option B
neglects the calorimeter constant. Options C and D stem from arithmetic signs or incorrect
temperature changes.
Question 3: During a lab evaluation of physical and chemical changes, a student encounters four
distinct clear, colorless solutions labeled W, X, Y, and Z. The following observations are noted:
1. Mixing W and X produces a sudden white precipitate that slowly dissolves upon adding excess
W.
2. Mixing W and Y causes a strong evolution of gas and a drop in temperature.
3. Mixing X and Z produces no visible precipitate, no gas evolution, and no measurable change in
temperature, but the resulting solution turns phenolphthalein dark pink.
Which of the following correctly classifies the nature of these changes?
A) 1 is a physical change; 2 and 3 are chemical changes.
B) 1 and 2 are chemical changes; 3 is a physical change due to lack of temperature change.
C) 1, 2, and 3 are all chemical changes.
, D) 1 and 3 are physical changes; 2 is a chemical change.
Correct Answer: C) 1, 2, and 3 are all chemical changes.
Explanation: Observation 1 involves precipitate formation and complexation chemistry
(dissolution in excess reagent), which involves the breaking and forming of chemical bonds
(chemical change). Observation 2 describes gas evolution and an endothermic chemical
reaction, which are clear signatures of a chemical change. Observation 3 indicates an acid-base
neutralization or pH shift that changes the chemical structure of the phenolphthalein indicator to
its basic pink form; despite the lack of dramatic macroscopic thermal or phase observations
initially, chemical species were altered to change the solution pH. Therefore, all three processes
represent chemical changes.
Question 4: A student measures the mass of an empty beaker five times and obtains the
following data: $25.321\text{ g}, 25.323\text{ g}, 25.320\text{ g}, 25.322\text{ g},$ and
$25.322\text{ g}$. The true mass of the beaker calibrated by a microbalance is $25.550\text{
g}$. Which of the following statements most accurately describes this set of data?
A) The data set has high precision and high accuracy.
B) The data set has low precision and high accuracy.
C) The data set has high precision and low accuracy, indicating a systematic error.
D) The data set has low precision and low accuracy, indicating a random error.
Correct Answer: C) The data set has high precision and low accuracy, indicating a
systematic error.
Explanation: The data values are closely grouped together (range of $0.003\text{ g}$),
indicating a high degree of reproducibility and precision. However, the average value ($\approx
25.322\text{ g}$) is significantly far from the true value ($25.550\text{ g}$), meaning the
accuracy is low. A consistent, directional deviation from the true value signifies a systematic
error, such as an improperly tared or miscalibrated balance. Random errors would result in a
wide scatter on both sides of the true value.
Question 5: A student is performing a liquid- liquid extraction to separate two immiscible
components. Liquid A has a density of $1.26\text{ g/mL}$ at $20^\circ\text{C}$ and Liquid B
has a density of $0.89\text{ g/mL}$ at $20^\circ\text{C}$. During the experiment, the
temperature of the room rises drastically to $45^\circ\text{C}$. If Liquid A expands significantly
more than Liquid B upon heating, what could potentially happen to the orientation of the layers
in the separatory funnel?
A) Liquid A will always remain the top layer because its mass remains constant.
Advanced Prep: Master Density, Specific
Heat, and Physical/Chemical Properties
Practice Questions & Detailed Explanations
Subject / Subtopic: General Chemistry I Lab (CHEM 121) / Density, Specific
Heat, and Physical vs. Chemical Properties
Question 1: A student is tasked with determining the density of an irregularly shaped, highly
porous ceramic material that reactively decomposes in water but remains completely inert in
cyclohexane ($\text{density} = 0.778\text{ g/mL}$). The dry mass of the ceramic sample is
$14.250\text{ g}$. To prevent the liquid from filling the internal pores during the displacement
measurement, the student coats the sample uniformly with a thin, water-insoluble polymer wax
($\text{density} = 0.920\text{ g/mL}$). The total mass of the coated ceramic sample is
$15.170\text{ g}$. When the coated sample is fully submerged in a graduated cylinder
containing cyclohexane, the liquid volume rises from $25.00\text{ mL}$ to $39.50\text{ mL}$.
Calculate the true density of the porous ceramic material itself.
A) $1.11\text{ g/mL}$
B) $1.06\text{ g/mL}$
C) $1.35\text{ g/mL}$
D) $0.982\text{ g/mL}$
Correct Answer: B) 1.06 g/mL
Explanation: To find the true density of the ceramic material, we must isolate its volume from the
volume of the polymer wax coating. First, determine the mass of the wax: $15.170\text{ g} -
14.250\text{ g} = 0.920\text{ g}$. Using the density of the wax, calculate its volume: $0.920\text{
g} / 0.920\text{ g/mL} = 1.000\text{ mL}$. Next, find the total displacement volume of the coated
sample: $39.50\text{ mL} - 25.00\text{ mL} = 14.50\text{ mL}$. Subtract the volume of the wax
to find the true volume of the ceramic material: $14.50\text{ mL} - 1.000\text{ mL} = 13.50\text{
mL}$. Finally, calculate the density of the ceramic: $14.250\text{ g} / 13.50\text{ mL} =
1.0556\text{ g/mL}$, which rounds to $1.06\text{ g/mL}$ based on significant figures. Option A
ignores the wax volume deduction. Option C miscalculates the mass-to-volume ratio, and Option
D fails to separate the mass and volume components accurately.
Question 2: In a calorimetry experiment designed to determine the specific heat capacity of an
unknown metal alloy, a $55.00\text{ g}$ sample of the alloy is heated to $99.50^\circ\text{C}$
and quickly transferred to a coffee-cup calorimeter containing $100.00\text{ g}$ of deionized
, water at $22.00^\circ\text{C}$. The highest equilibrium temperature reached by the system is
recorded as $26.80^\circ\text{C}$. If the calorimeter constant (heat capacity of the calorimeter
itself) is $12.5\text{ J/}^\circ\text{C}$, and the specific heat of water is $4.184\text{
J/g}^\circ\text{C}$, what is the specific heat capacity of the metal alloy?
A) $0.518\text{ J/g}^\circ\text{C}$
B) $0.473\text{ J/g}^\circ\text{C}$
C) $0.442\text{ J/g}^\circ\text{C}$
D) $0.534\text{ J/g}^\circ\text{C}$
Correct Answer: A) 0.518 J/g°C
Explanation: The heat lost by the metal alloy must equal the heat gained by both the water and
the calorimeter: $-q_{\text{metal}} = q_{\text{water}} + q_{\text{calorimeter}}$. Calculate heat
gained by water: $q_{\text{water}} = m \cdot c \cdot \Delta T = 100.00\text{ g} \cdot 4.184\text{
J/g}^\circ\text{C} \cdot (26.80^\circ\text{C} - 22.00^\circ\text{C}) = 100.00 \cdot 4.184 \cdot
4.80 = 2008.32\text{ J}$. Calculate heat gained by the calorimeter: $q_{\text{calorimeter}} =
C_{\text{cal}} \cdot \Delta T = 12.5\text{ J/}^\circ\text{C} \cdot 4.80^\circ\text{C} = 60.0\text{
J}$. Total heat gained = $2008.32\text{ J} + 60.0\text{ J} = 2068.32\text{ J}$. Therefore,
$q_{\text{metal}} = -2068.32\text{ J}$. Set up the equation for the metal: $-2068.32\text{ J} =
55.00\text{ g} \cdot c_{\text{metal}} \cdot (26.80^\circ\text{C} - 99.50^\circ\text{C})$. Thus, $-
2068.32 = 55.00 \cdot c_{\text{metal}} \cdot (-72.70)$, yielding $c_{\text{metal}} = 2068.32 /
3998.5 = 0.51727\text{ J/g}^\circ\text{C}$, or $0.518\text{ J/g}^\circ\text{C}$. Option B
neglects the calorimeter constant. Options C and D stem from arithmetic signs or incorrect
temperature changes.
Question 3: During a lab evaluation of physical and chemical changes, a student encounters four
distinct clear, colorless solutions labeled W, X, Y, and Z. The following observations are noted:
1. Mixing W and X produces a sudden white precipitate that slowly dissolves upon adding excess
W.
2. Mixing W and Y causes a strong evolution of gas and a drop in temperature.
3. Mixing X and Z produces no visible precipitate, no gas evolution, and no measurable change in
temperature, but the resulting solution turns phenolphthalein dark pink.
Which of the following correctly classifies the nature of these changes?
A) 1 is a physical change; 2 and 3 are chemical changes.
B) 1 and 2 are chemical changes; 3 is a physical change due to lack of temperature change.
C) 1, 2, and 3 are all chemical changes.
, D) 1 and 3 are physical changes; 2 is a chemical change.
Correct Answer: C) 1, 2, and 3 are all chemical changes.
Explanation: Observation 1 involves precipitate formation and complexation chemistry
(dissolution in excess reagent), which involves the breaking and forming of chemical bonds
(chemical change). Observation 2 describes gas evolution and an endothermic chemical
reaction, which are clear signatures of a chemical change. Observation 3 indicates an acid-base
neutralization or pH shift that changes the chemical structure of the phenolphthalein indicator to
its basic pink form; despite the lack of dramatic macroscopic thermal or phase observations
initially, chemical species were altered to change the solution pH. Therefore, all three processes
represent chemical changes.
Question 4: A student measures the mass of an empty beaker five times and obtains the
following data: $25.321\text{ g}, 25.323\text{ g}, 25.320\text{ g}, 25.322\text{ g},$ and
$25.322\text{ g}$. The true mass of the beaker calibrated by a microbalance is $25.550\text{
g}$. Which of the following statements most accurately describes this set of data?
A) The data set has high precision and high accuracy.
B) The data set has low precision and high accuracy.
C) The data set has high precision and low accuracy, indicating a systematic error.
D) The data set has low precision and low accuracy, indicating a random error.
Correct Answer: C) The data set has high precision and low accuracy, indicating a
systematic error.
Explanation: The data values are closely grouped together (range of $0.003\text{ g}$),
indicating a high degree of reproducibility and precision. However, the average value ($\approx
25.322\text{ g}$) is significantly far from the true value ($25.550\text{ g}$), meaning the
accuracy is low. A consistent, directional deviation from the true value signifies a systematic
error, such as an improperly tared or miscalibrated balance. Random errors would result in a
wide scatter on both sides of the true value.
Question 5: A student is performing a liquid- liquid extraction to separate two immiscible
components. Liquid A has a density of $1.26\text{ g/mL}$ at $20^\circ\text{C}$ and Liquid B
has a density of $0.89\text{ g/mL}$ at $20^\circ\text{C}$. During the experiment, the
temperature of the room rises drastically to $45^\circ\text{C}$. If Liquid A expands significantly
more than Liquid B upon heating, what could potentially happen to the orientation of the layers
in the separatory funnel?
A) Liquid A will always remain the top layer because its mass remains constant.
