# AP Chemistry: 1.1 Moles and Molar Mass – Exam Style questions with Answer- MCQ

Questions

Which of the following is the net ionic equation for the overall reaction that occurs as the cell operates?

(A) $$Ag(s) + Zn^{2+}(aq) \rightarrow Ag^{2+}(aq) + Zn(s)$$
(B)  $$Ag^{+}(aq) + Zn^{2+}(aq) + 3 e \rightarrow AgZn(s)$$
(C) $$Ag^{+}(aq) + Zn(s) \rightarrow Ag(s) + Zn^{2+}$$(aq) + e^{-}\)
(D) $$2 Ag^{+}(aq) + Zn(s) \rightarrow 2 Ag(s) + Zn^{2+}$$(aq)

Ans: D

To determine the net ionic equation for the overall reaction, we need to consider the two half-reactions and their standard reduction potentials given in the table:

Agt(aq) + e -+ Ag(s) E° = +0.80 V Zn2+(aq) + 2 e- – Zn(s) E° = -0.76 V

The half-reaction with the more positive reduction potential (Ag/Ag) will be the reduction reaction, while the half-reaction with the less positive (or more negative) reduction potential (Zn2+/Zn) will be the
oxidation reaction.

The overall reaction is obtained by multiplying the two half-reactions to cancel out the electrons transferred and then adding them:

Reduction: Ag*(aq) + e — > Ag(s) Oxidation: Zn(s) -> Zn2+(ag) + 2 e-

Overall reaction: Agt(aq) + Zn(s) -> Ag(s) + Zn2+(aq)

Therefore, the correct answer is (C): Ag+(aq) + Zn(s) -> Ag(s) + Zn2+(aq) + e-

Questions

Another sample of eggshell reacts completely with 4.0 mL of an HCl(aq) solution of unknown concentration. If the reaction produced 0.095 atm of gas, the concentration of the HCl(aq) solution was at least
(A) 0.0020 M
(B) 0.050 M
(C) 0.50 M
(D) 1.0 M

Ans: D

$$P=0.095$$ atm $$\rightarrow$$ corresponds to $$0.20 \mathrm{~g} \mathrm{CaCO}_3$$
So, moles of $$\mathrm{CaCO}_3=\frac{0.20 \mathrm{~g}}{100 \mathrm{gm}{mol}^{-1}}=0.002 \mathrm{moles}$$
moles of $$\mathrm{HCl}=\left(\right.$$ moles of $$\left.\mathrm{CaCO}_3 \times 2\right) \Rightarrow 0.002 \times 2$$
So, molarity of $$\mathrm{HCl}=\frac{\text { moles }}{\text { Volume in } \mathrm{L}} \Rightarrow \frac{0.004}{4} \times 1000$$
Concentration of $$\mathrm{HCl}=1 \mathrm{~m} \rightarrow$$ option $$(\mathrm{D})$$

Questions

The mass percent of $$CaCO_{3}$$(s) in the eggshell  is closest to
(A) 30%
(B) 45%
(C) 60%
(D) 75%

Ans: D

(36) Pressure of gas $$=0.870-0.800=0.07 \mathrm{~atm}$$ Mass of $$\mathrm{CaCO}_3$$ corresponding to it $$\Rightarrow 0.15 \mathrm{~g}$$ (approx)
So, $$\% \mathrm{CaCO}_3=\frac{0.15}{0.20} \times 100 \Rightarrow 75 \%$$

Questions

$$CO_2(g) + 2 LiOH(s) → Li_2CO3(aq) + H_2O$$(l)

In a one-person spacecraft, an astronaut exhales 880 g of $$CO_2$$(g) (molar mass 44 g/mol) per day. To prevent the buildup of $$CO_{2}$$(g) in the spacecraft, a device containing LiOH(s) is used to remove the $$CO_2$$(g), as represented by the equation above. What mass of LiOH(s) (molar mass 24 g/mol) is needed to react with all of the $$CO_2$$(g) produced by an astronaut in one day?
(A) 40. g
(B) 240 g
(C) 480 g
(D) 960 g

Ans: D

First, let’s calculate the moles of $$CO_2$$ produced by the astronaut in one day:

Mass of$$CO_2$$ produced per day = 880 g
Molar mass of $$CO_2$$ =44 g/mol

$\text{moles of } CO_2 =\frac{\text{mass of } CO_2}{\text{molar mass of } CO_2}$

$\text{moles of } CO_2 = \frac{880 \, \text{g}}{44 \, \text{g/mol}} = 20 \, \text{mol}$

According to the balanced chemical equation, 1 mole of $$CO_2$$ reacts with 2 moles of $$LiQH$$. So, to find the moles of $$LiQH$$ needed, we multiply the moles of $$CO_2$$ by the stoichiometric coefficient ratio:

$\text{moles of } LiQH = 2 \times \text{moles of } CO_2$

$\text{moles of } LiQH = 2 |times 20 , \text{mol} = 40 \, \text{mol}$

Now, let’s calculate the mass of $$LiQH$$ needed:

Molar mass of $$LiQH$$ = 24 g/mol

$\text{mass of } LiQH = \text{moles of } LiQH \times \text{molar mass of } LiQH$

$\text{mass of } LiQH = 40 \, \text{mol} \times 24 \, \text{g/mol} = 960 \, \text{g}$

So, the mass of $$LiQH$$ needed to react with all of the $$CO_2$$ produced by the astronaut in one day is 960 g.

Therefore, the correct answer is (D) 960 g.

### Question

A group of students was asked to recover Cu(s) from a blue-green aqueous solution containing an unknown concentration of $$Cu^{2+}$$(aq). The students took a 100.0 mL sample of the solution and added an excess of 1.0 $$M Na_3PO_4$$(aq), causing the$$Cu^{2+}$$(aq) to precipitate as Cu3(PO4)2(s), as shown in step 1 below.

If 3.8 g of $$Cu_{3}(PO_{4}$$)2(s) was recovered from  step 1, what was the approximate $$[Cu^{2+}$$] in the original solution? (The molar mass of$$Cu_{3}(PO_{4})_{2}$$ is 381 g/mol.)
(A) 0.10 M
(B) 0.30 M
(C) 1.0 M
(D) 3.0 M

Ans:B

To find the concentration of $$Cu^{2+}$$ in the original solution, we need to consider the stoichiometry of the reactions involved.

1. From step 1, we know that 3 moles of $$Cu^{2+}$$ ions are produced for every mole of $$Cu_{3}(PO_{4})_{2}$$ formed.

2. The molar mass of $$Cu_{3}(PO_{4})_{2}$$ is 381 g/mol.

Given that 3.8 g of $$Cu_{3}(PO_{4})_{2}$$ was recovered, we can calculate the number of moles of $$Cu_{3}(PO_{4})_{2}$$ using its molar mass:

$\text{Moles of } Cu_{3}(PO_{4})_{2} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{3.8 \text{ g}}{381 \text{ g/mol}} \approx 0.01 \text{ mol}$

Since 3 moles of $$Cu^{2+}$$ ions are produced for every mole of $$Cu_{3}(PO_{4})_{2}$$ formed, the number of moles of $$Cu^{2+}$$ ions is three times the number of moles of $$Cu_{3}(PO_{4})_{2}$$:

$\text{Moles of } Cu^{2+} = 3 \times 0.01 \text{ mol} = 0.03 \text{ mol}$

Next, we need to calculate the volume of the original solution:

$\text{Volume of original solution} = \frac{\text{Volume of solution after step 1}}{\text{Dilution factor}}$

The dilution factor is the ratio of the volume of the original solution to the volume of the sample taken (100.0 mL). Since the sample was taken from the original solution, the dilution factor is 1.

Now, we can use the concentration formula:

$\text{Concentration} = \frac{\text{Moles}}{\text{Volume}}$

Given that the volume is in liters, we need to convert the volume of the original solution to liters (divide by 1000).

$\text{Volume of original solution} = \frac{100.0 \text{ mL}}{1000} = 0.100 \text{ L}$

Now we can calculate the concentration:

$\text{Concentration of } Cu^{2+} = \frac{0.03 \text{ mol}}{0.100 \text{ L}} = 0.30 \text{ M}$

So, the approximate $$[Cu^{2+}]$$ in the original solution is 0.30 M, which corresponds to option (B).

### Question

A group of students was asked to recover Cu(s) from a blue-green aqueous solution containing an unknown concentration of $$Cu^{2+}$$(aq). The students took a 100.0 mL sample of the solution and added an excess of 1.0 $$M Na_3PO_4$$(aq), causing the$$Cu^{2+}$$(aq) to precipitate as Cu3(PO4)2(s), as shown in step 1 below.

The mass of the Cu(s) produced in step 3 was slightly more than the mass predicted from the  3.8 g of$$Cu_{3}(PO_{4})2(s)$$ recovered from step 1. Which of the following could account for the  discrepancy in the yield of Cu(s) from step 3 ?
(A) All of the $$Cu_{3}(PO_{4})$$2(s) dissolved in the HCl(aq).
(B) Some of the Cu(s) adhered to the side of the funnel used to filter the solution.
(C) Some unreacted Zn(s) was mixed in with the Cu(s).
(D) Too much $$Na_{3}PO_{4}$$(aq) was added to the original$$Cu^{2+}$$(aq) solution.

Ans:C

The mass of Cu(s) produced in step 3 being slightly more than the mass predicted from the 3.8 g of $$Cu_{3}(PO_{4})_{2(s)}$$ recovered from step 1 suggests that there might have been some additional factors influencing the yield of Cu(s) in step 3.

Let’s consider each option:

(A) All of the $$Cu_{3}(PO_{4})_{2(s)}$$ dissolved in the HCl(aq):
If all of the $$Cu_{3}(PO_{4})_{2(s)}$$ dissolved in the HCl(aq), there would not be any solid precipitate left to react with Zn(s) to form Cu(s). This would lead to a lower yield of Cu(s) rather than a higher yield. Therefore, this option is not likely to account for the discrepancy.

(B) Some of the Cu(s) adhered to the side of the funnel used to filter the solution:
If some of the Cu(s) adhered to the side of the funnel, it would not contribute to the measured yield of Cu(s) in the final step, leading to a lower yield. Therefore, this option is not likely to account for the higher yield observed.

(C) Some unreacted Zn(s) was mixed in with the Cu(s):
If some unreacted Zn(s) was mixed in with the Cu(s), it would contribute to the measured mass of Cu(s), leading to a higher yield. This option could potentially account for the discrepancy observed.

(D) Too much $$Na_{3}PO_{4(aq)}$$ was added to the original $$Cu^{2+}(aq)$$ solution:
Adding too much $$Na_{3}PO_{4(aq)}$$ could lead to excess $$Na_{3}PO_{4(aq)}$$ in the solution, which might interfere with the subsequent reactions and affect the yield of Cu(s). However, this is less likely to directly lead to a higher yield of Cu(s).

Therefore, option (C) “Some unreacted Zn(s) was mixed in with the Cu(s)” is the most plausible explanation for the discrepancy in the yield of Cu(s) from step 3.

### Question

$\mathrm{Li}_3 \mathrm{~N}(s)+2 \mathrm{H}_2(g) \rightleftarrows \mathrm{LiNH}_2(s)+2 \mathrm{LiH}(s) \quad \Delta H^{\circ}=-192 \mathrm{~kJ} / \mathrm{mol}_{r x n}$

Because pure $$\mathrm{H}_2$$ is a hazardous substance, safer and more cost effective techniques to store it as a solid for shipping purposes have been developed. One such method is the reaction represented above, which occurs at $$200^{\circ} \mathrm{C}$$.

When 70. g of$$Li_3$$N(s) (molar mass 35 g/mol)  reacts with excess H2(g), 8.0 g of $$LiH(s)$$ is produced. The percent yield is closest to
(A) 17%
(B) 25%
(C) 50.%
(D) 100%

Ans:B

To find the percent yield of $$\mathrm{LiH}(s)$$, we first need to determine the theoretical yield of $$\mathrm{LiH}(s)$$ from the given reaction and then compare it to the actual yield.

1. Calculate the number of moles of $$\mathrm{Li}_3\mathrm{N}(s)$$ used:
$\text{Moles of } \mathrm{Li}_3\mathrm{N}(s) = \frac{\text{mass}}{\text{molar mass}} = \frac{70 \, \text{g}}{35 \, \text{g/mol}} = 2 \, \text{mol}$

2. Using the stoichiometry of the reaction, determine the number of moles of $$\mathrm{LiH}(s)$$ produced:
From the balanced equation, we see that the coefficient of $$\mathrm{LiH}(s)$$ is 2. This means that for every 1 mole of $$\mathrm{Li}_3\mathrm{N}(s)$$ that reacts, 2 moles of $$\mathrm{LiH}(s)$$ are produced.

So, moles of $$\mathrm{LiH}(s)$$ produced = $$2 \times \text{moles of } \mathrm{Li}_3\mathrm{N}(s) = 2 \times 2 \, \text{mol} = 4 \, \text{mol}$$

3. Calculate the theoretical yield of $$\mathrm{LiH}(s)$$:
$\text{Theoretical yield} = \text{moles of } \mathrm{LiH}(s) \times \text{molar mass of } \mathrm{LiH}(s)$
$\text{Theoretical yield} = 4 \, \text{mol} \times \frac{6.94 + 1.008}{1} \, \text{g/mol} = 4 \times 7.948 = 31.792 \, \text{g}$

4. Calculate the percent yield:
$\text{Percent yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\%$
$\text{Percent yield} = \left( \frac{8.0 \, \text{g}}{31.792 \, \text{g}} \right) \times 100\% \approx 25\%$

So, the closest answer choice is: (B) 25%

Questions

If a pure sample of an oxide of sulfur contains 40. percent sulfur and 60. percent oxygen by mass, then the empirical formula of the oxide is
(A) $$SO_{3}$$
(B) $$SO_{4}$$
(C) $$S_{2}O_{6}$$
(D) $$S_{2}O_{8}$$

Ans: A

To determine the empirical formula of the oxide of sulfur, we need to find the simplest whole-number ratio of sulfur to oxygen atoms in the compound.

Given that the oxide contains 40% sulfur and 60% oxygen by mass, we can assume we have 100 g of the compound for simplicity.

So, for 100 g of the compound:
Mass of sulfur = 40 g
Mass of oxygen = 60 g

Number of moles of sulfur:
$\text{Moles of sulfur} = \frac{\text{Mass of sulfur}}{\text{Molar mass of sulfur}} = \frac{40 \, \text{g}}{32.07 \, \text{g/mol}} \approx 1.247 \, \text{mol}$

Number of moles of oxygen:
$\text{Moles of oxygen} = \frac{\text{Mass of oxygen}}{\text{Molar mass of oxygen}} = \frac{60 \, \text{g}}{16.00 \, \text{g/mol}} = 3.75 \, \text{mol}$

Now, we need to find the simplest whole-number ratio of sulfur to oxygen atoms by dividing the number of moles of each element by the smallest number of moles:

Dividing by 1.247 (the number of moles of sulfur), we get approximately:
Sulfur: $$1$$ (rounded to the nearest whole number)
Oxygen: $$3$$ (rounded to the nearest whole number)

So, the empirical formula of the oxide is $$SO_3$$, as there is one sulfur atom for every three oxygen atoms.

Questions

$$Cu_{(s)}+4HNO_{3}(aq)\rightarrow Cu(No_{3})_{2}(aq)+2No_{2(g)}+2H_{2}o_{(l)}$$

Each student in a class placed a $$2.00 \mathrm{~g}$$ sample of a mixture of $$\mathrm{Cu}$$ and $$\mathrm{Al}$$ in a beaker and placed the beaker in a fume hood. The students slowly poured $$15.0 \mathrm{~mL}$$ of $$15.8 \mathrm{M} \mathrm{HNO}_3(a q)$$ into their beakers. The reaction between the copper in the mixture and the $$\mathrm{HNO}_3(a q)$$ is represented by the equation above. The students observed that a brown gas was released from the beakers and that the solutions turned blue, indicating the formation of $$\mathrm{Cu}^{2+}(a q)$$. The solutions were then diluted with distilled water to known volumes.

The students determined that the reaction produced 0.010 mol of $$Cu(No_{3})_{2}$$. Based on the measurement, what was the percent of Cu by mass in the original 2.00 g sample of the mixture?
(A) 16%
(B) 32%
(C) 64%
(D) 96%

Ans: B

Given:
The balanced chemical equation for the reaction is: $$\text{Cu(s)} + 4\text{HNO}_3(\text{aq}) \rightarrow \text{Cu(NO}_3\text{)}_2(\text{aq}) + 2\text{NO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})$$
The students determined that the reaction produced 0.010 mol of $$\text{Cu(NO}_3\text{)}_2$$.
The original sample of the mixture weighed 2.00 g.

Step 1: Find the moles of copper that reacted to produce 0.010 mol of $$\text{Cu(NO}_3\text{)}_2$$.
From the balanced chemical equation, we can see that 1 mol of Cu reacts to produce 1 mol of $$\text{Cu(NO}_3\text{)}_2$$.
Therefore, 0.010 mol of $$\text{Cu(NO}_3\text{)}_2$$ was produced from 0.010 mol of Cu.

Step 2: Calculate the mass of copper present in the original sample.
Molar mass of Cu = 63.55 g/mol
Mass of Cu = Moles of Cu × Molar mass of Cu
Mass of Cu = 0.010 mol × 63.55 g/mol = 0.6355 g

Step 3: Calculate the percentage of copper by mass in the original 2.00 g sample of the mixture.
Percentage of Cu by mass = $$\left( \frac{\text{Mass of Cu}}{\text{Total mass of the mixture}} \right) \times 100\%$$
Percentage of Cu by mass = $$\left( \frac{0.6355 \text{ g}}{2.00 \text{ g}} \right) \times 100\%$$
Percentage of Cu by mass = 31.775%

Therefore, the correct answer is (B) 32%.

Questions

Based on the structures shown above, which of the following statements identifies the compound with the higher boiling point and provides the best explanation for the higher boiling point?
(A) Compound 1, because it has stronger dipole-dipole forces than compound 2
(B) Compound 1, because it forms hydrogen bonds, whereas compound 2 does not
(C) Compound 2, because it is less polarizable and has weaker London dispersion forces than compound 1
(D) Compound 2, because it forms hydrogen bonds, whereas compound 1 does not

Ans: D

To determine which compound has the higher boiling point and the best explanation, we need to consider the intermolecular forces present in each compound.

(A) Compound 1, because it has stronger dipole-dipole forces than compound 2 This statement is incorrect. While trimethylamine (Compound 1) has a permanent dipole due to the electronegativity difference between nitrogen and hydrogen, n-butylamine (Compound 2) also has a permanent dipole due to the N-H bond. The dipole-dipole forces are likely stronger in Compound 2 due to the presence of the polar N-H bond.

(B) Compound 1, because it forms hydrogen bonds, whereas compound 2 does not This statement is incorrect. Trimethylamine (Compound 1) does not have any hydrogen atoms attached to the highly electronegative nitrogen atom, so it cannot form hydrogen bonds. However, n-butylamine (Compound 2) has an N-H bond, allowing it to form hydrogen bonds.

(C) Compound 2, because it is less polarizable and has weaker London dispersion forces than compound 1 This statement is incorrect. London dispersion forces (van der Waals forces) are present in both compounds due to their non-polar hydrocarbon portions. However, n-butylamine (Compound 2) is likely more polarizable due to the presence of the polar N-H bond, leading to stronger dispersion forces compared to trimethylamine (Compound 1).

(D) Compound 2, because it forms hydrogen bonds, whereas compound 1 does not This statement is correct. n-Butylamine (Compound 2) has an N-H bond, which can form hydrogen bonds with other molecules or with itself. Hydrogen bonding is a strong intermolecular force that requires more energy to overcome, resulting in a higher boiling point. On the other hand, trimethylamine (Compound 1) does not have any hydrogen atoms attached to the nitrogen atom, so it cannot form hydrogen bonds.

Therefore, the correct answer is (D) Compound 2, because it forms hydrogen bonds, whereas compound 1 does not.

### Question

Which of the following numerical expressions gives the number of moles in 5.0g of CaO?

A $$5.0g\times 56g/mol$$

B$$\frac{5.0g}{56g/mol}$$

C$$\frac{56g/mol}{5.0g}$$

D$$\frac{1}{5.0g}\times \frac{1}{50g/mol}$$

Ans:B

Multiplying the mass of the sample by the molar mass of CaO gives an irrelevant value and unit. Dividing the mass of the sample by the molar mass of CaO would give the number of moles of CaO in the sample.

### Question

In a lab, a student is given a 21g sample of pure Cu metal. Which of the following pieces of information is most useful for determining the number of Cu atoms in the sample? Assume that the pressure and temperature in the lab are 1.0atm  and 25°C?

1. The molar mass of Cu.
2. The density of Cu at 25°C.
3. The volume of the Cu sample.
4. The ratio of the two main isotopes found in pure Cu.

Ans-A

The density could be used with the given mass to determine the volume of the Cu sample, but this would not be helpful for determining the number of Cu atoms in the sample. However, by dividing the mass of the sample by the molar mass of Cu, the number of moles of Cu in the sample can be calculated. A simple multiplication of that number with Avogadro’s number will provide the desired information.

### Question

1.0 mol sample of which of the following compounds has the greatest mass?

A $$NO$$

B. $$NO_{2}$$

C $$N_{2}O$$

D $$N_{2}O_{5}$$

Ans:D

The mass of 1.0mol of NO is less than the mass of 1.0mol of N2O5 because it has fewer N and O atoms and thus a smaller molar mass.

### Question

$$C_3H_8(g) + 4 Cl_2(g) → C_3H_4Cl(g) + 4 HCl(g)$$

A 6.0 mol sample of$$C_{3}H_8(g)$$ and a 20. mol sample of $$Cl_{2}$$(g) are placed in a previously evacuated vessel, where they react according to the equation above. After one of the reactants has been totally consumed, how many moles of HCl(g) have been produced?

(A) 4.0 mo

(B) 8.0 mol

(C) 20. mol

(D) 24 mol

Ans:C

### Question

$$CO_2(g) + 2 LiOH(s) → Li_2CO3(aq) + H_2O$$(l)

In a one-person spacecraft, an astronaut exhales 880 g of $$CO_2$$(g) (molar mass 44 g/mol) per day. To prevent the buildup of $$CO_{2}$$(g) in the spacecraft, a device containing LiOH(s) is used to remove the $$CO_2$$(g), as represented by the equation above. What mass of LiOH(s) (molar mass 24 g/mol) is needed to react with all of the $$CO_2$$(g) produced by an astronaut in one day?
(A) 40. g
(B) 240 g
(C) 480 g
(D) 960 g