ACS General Chemistry
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ACS General Chemistry is a trademark of its owner, which is not affiliated with and does not endorse Test Treasure Publication or TTP Prep.
10 sample questions for the ACS General Chemistry, with the correct answer and an explanation for each. No account needed. Try each one before you open the answer.
Sections covered on the exam: Atoms, Lightandatomicspectra, Periodictrends, Properties Of Matter, Statesofmatterandgases, Phasechanges, Solutions, Bonding And Intermolecular Interactions, Ionicandcovalentbonding, Molecularshapeandpolarity, Intermolecularforces, Reactions, Reactiontypes, Stoichiometry, Solutionstoichiometry, Kinetics And Equilibrium, Kinetics, Reactionmechanisms, Safety, Math, and Data in the Laboratory, So₃, Acids And Bases, Phandbuffers, Titrationsandsolubility, Thermodynamics, Electrochemistry, Redoxandelectrochemicalcells, Nuclear Chemistry, Nuclearchemistry, Safety, Math, And Data In The Laboratory, Laboratorymathanddata, Δs° (J/K).
Question 1 · Atoms
Naturally occurring gallium consists of two isotopes, ⁶⁹Ga (68.926 amu) and ⁷¹Ga (70.925 amu). The atomic mass of gallium is 69.723 amu. What is the percent natural abundance of ⁶⁹Ga?
Answer: B. 60.1%
Let x be the fraction of ⁶⁹Ga: 68.926x + 70.925(1 − x) = 69.723, so x = (70.925 − 69.723)/(70.925 − 68.926) = 1.202/1.999 = 0.601, or 60.1%. The value 39.9% is the abundance of ⁷¹Ga, obtained by solving for the wrong isotope. Assuming 50.0% ignores the weighting, and 69.7% confuses the average mass with a percentage.
Question 2 · Solutions
A 6.00 M aqueous solution of H₂SO₄ (98.08 g/mol) has a density of 1.338 g/mL. What is the molality of H₂SO₄ in this solution?
Answer: C. 8.01 m
One liter of solution has a mass of 1338 g and contains 6.00 mol × 98.08 g/mol = 588.5 g of H₂SO₄, so the mass of water is 749.5 g and molality = 6.00 mol/0.7495 kg = 8.01 m. The value 6.00 m assumes molarity equals molality, and 4.48 m divides by the mass of the whole solution instead of the solvent. The value 14.6 m assumes a solution mass of 1000 g, ignoring the density.
Question 3 · Stoichiometry
Ammonia reacts with oxygen according to 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g). A mixture of 20.0 g of NH₃ and 40.0 g of O₂ reacts completely. What mass of which reactant remains?
Answer: D. 2.97 g of NH₃
Moles: NH₃ = 20.0/17.03 = 1.174 mol and O₂ = 40.0/32.00 = 1.250 mol. Consuming all the NH₃ would need 1.174 × 5/4 = 1.468 mol O₂, which is more than available, so O₂ is limiting. O₂ consumes 1.250 × 4/5 = 1.000 mol NH₃ (17.03 g), leaving 20.0 − 17.03 = 2.97 g NH₃. The 6.98 g O₂ value comes from wrongly treating NH₃ as limiting and subtracting 40.0 g from the 46.98 g of O₂ it would require.
Question 4 · So₃
N₂O₄(g) ⇌ 2NO₂(g) has ΔH° = +57 kJ. Which change to the equilibrium mixture increases both the amount of NO₂ present and the value of K?
Answer: B. Raising the temperature
Only a temperature change alters K; because the forward reaction is endothermic, raising the temperature increases K and shifts the equilibrium toward NO₂. Adding N₂O₄ and increasing the volume (favoring the side with more moles of gas) both increase the amount of NO₂, but K stays the same. A catalyst changes neither K nor the equilibrium composition.
Question 5 · Safety, Math, and Data in the Laboratory
Ammonia is oxidized in the first step of nitric acid production: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g). −46.1 0 +90.3 −241.8 How much heat is released when 10.0 g of NH₃ reacts completely with excess O₂?
Answer: A. 133 kJ
= 0.587 mol × (905.2 kJ / 4 mol) = 133 kJ. The value 532 kJ ignores the coefficient of 4 on NH₃, 187 kJ adds the NH₃ term instead of subtracting it, and 160 kJ omits the reactant term entirely.
Question 6 · Laboratorymathanddata
Four students each determine the density of a liquid whose accepted density is 1.20 g/mL. Their three trials are shown. W 1.20 1.35 1.05 X 1.31 1.32 1.31 Y 1.19 1.20 1.21 Z 1.02 1.40 1.28 Which student's data are precise but inaccurate, suggesting a systematic error?
Answer: B. Student X
Student X's trials agree within 0.01 g/mL (high precision) but average about 1.31 g/mL, about 9% above the accepted value, a consistent offset typical of a systematic error such as a miscalibrated pipet. Student W's trials average exactly 1.20 g/mL but scatter widely (accurate on average, imprecise), Student Y's data are both accurate and precise, and Student Z's data are neither.
Question 7 · Phasechanges
On the phase diagram of water, the solid–liquid equilibrium line slopes to the left (negative slope). Which statement best explains this feature?
Answer: A. Ice is less dense than liquid water, so higher pressure favors the liquid.
Because ice occupies a larger volume than the same mass of liquid, increasing pressure shifts the equilibrium toward the denser liquid, lowering the melting point as pressure rises. If ice were denser than the liquid, the line would slope to the right, as it does for most substances. The triple point location does not determine the slope, and the explanation does not depend on hydrogen bonds strengthening with pressure.
Question 8 · Reactions
Using the activity series, which metal reacts with aqueous Pb(NO₃)₂ but does NOT react with aqueous Zn(NO₃)₂?
Answer: A. Fe
A metal displaces the cation of any metal below it in the activity series (… Mg, Al, Zn, Fe, Ni, Sn, Pb, H, Cu, Ag …). Iron lies above lead but below zinc, so Fe reduces Pb²⁺ to Pb yet cannot reduce Zn²⁺. Magnesium is above both and reacts with both solutions, while copper and silver are below both and react with neither.
Question 9 · Safety, Math, and Data in the Laboratory
Pure N₂O₄ at an initial pressure of 1.00 atm is sealed in a rigid container and establishes the equilibrium N₂O₄(g) ⇌ 2NO₂(g) at constant temperature. The total pressure at equilibrium is 1.40 atm. What is Kp?
Answer: C. 1.07
If N₂O₄ drops by x, NO₂ rises by 2x, and the total pressure is 1.00 + x = 1.40 atm, so x = 0.40 atm. Then P(N₂O₄) = 0.60 atm, P(NO₂) = 0.80 atm, and Kp = 0.80²/0.60 = 1.07. The value 0.267 sets P(NO₂) = x instead of 2x, 1.33 forgets to square P(NO₂), and 0.640 divides by the initial rather than the equilibrium N₂O₄ pressure.
Question 10 · Thermodynamics
The standard enthalpy of formation of methanol, CH₃OH(l), is −238.6 kJ/mol. For which equation is ΔH° equal to ΔHf° of CH₃OH(l)?
Answer: B. C(graphite) + 2H₂(g) + ½O₂(g) → CH₃OH(l)
The formation reaction produces exactly one mole of the compound from its elements in their standard states. Graphite, H₂(g), and O₂(g) are the standard states, and the product is 1 mol CH₃OH(l). Diamond is not the standard state of carbon, the doubled equation has
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