Avogadro's number Transitional Reasoning Practice

Chemistry · The Mole & Stoichiometry · Chemistry · The Mole

5 Chemistry practice questions on Avogadro's number. Questions that build towards applied reasoning.

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Questions

  1. 1.

    In an investigation modeled after the historical determination of fundamental physical constants, a radioactive emitter releases alpha particles at a constant rate of 3.01×10113.01 \times 10^{11} particles per second into a collection chamber. Each alpha particle gains two electrons upon stopping, converting into a neutral helium atom. After a continuous collection period of 2.00×1072.00 \times 10^7 seconds, the mass of the accumulated helium gas (He\text{He}, molar mass = 4.00 g/mol4.00\text{ g/mol}) is determined to be 4.00×10−5 g4.00 \times 10^{-5}\text{ g}. Based on these experimental data, what is the calculated value of Avogadro's number?

    • A) 6.02×1023 mol−16.02 \times 10^{23}\text{ mol}^{-1}
    • B) 3.01×1023 mol−13.01 \times 10^{23}\text{ mol}^{-1}
    • C) 1.51×1023 mol−11.51 \times 10^{23}\text{ mol}^{-1}
    • D) 1.20×1024 mol−11.20 \times 10^{24}\text{ mol}^{-1}
  2. 2.

    A deep-space probe uses an electrostatic ion thruster that ionizes gaseous xenon (Xe\text{Xe}, molar mass = 131 g/mol) into singly charged cations and expels them at high velocity. Ground telemetry confirms that the propulsion system expels a steady stream of 2.0×10182.0 \times 10^{18} xenon particles per second. If the thruster operates continuously at this rate for exactly 300 minutes, what total mass of xenon propellant is consumed during the maneuver? (Take Avogadro's number as NA=6.0×1023 mol−1N_A = 6.0 \times 10^{23}\text{ mol}^{-1}.)

    • A) 0.131 g
    • B) 0.786 g
    • C) 78.6 g
    • D) 7.86 g
  3. 3.

    An analytical chemist dissolves 0.050 mol of solid ammonium sulfate, (NH4)2SO4(NH_4)_2SO_4, completely in distilled water to prepare 500 mL of a homogeneous stock solution. Exactly 50.0 mL of this solution is transferred to an Erlenmeyer flask for analysis. Taking Avogadro's number as NA=6.02×1023 mol−1N_A = 6.02 \times 10^{23} \text{ mol}^{-1}, what is the total number of dissolved ions present in the transferred sample?

    • A) 3.01×10213.01 \times 10^{21} ions
    • B) 6.02×10216.02 \times 10^{21} ions
    • C) 9.03×10229.03 \times 10^{22} ions
    • D) 9.03×10219.03 \times 10^{21} ions
  4. 4.

    A mass spectrometer measures the absolute mass of a single molecule of a homonuclear diatomic gas, X2X_2, to be 4.65×10−23 g4.65 \times 10^{-23}\text{ g}. A chemist collects a 14.0 g14.0\text{ g} sample of this pure gaseous substance to study its reactivity. Taking Avogadro's number as 6.02×1023 mol−16.02 \times 10^{23}\text{ mol}^{-1}, which choice correctly identifies element XX and gives the total number of individual atoms contained in the 14.0 g14.0\text{ g} sample?

    • A) Silicon (Si\text{Si}), and it contains 3.01×10233.01 \times 10^{23} atoms
    • B) Nitrogen (N\text{N}), and it contains 6.02×10236.02 \times 10^{23} atoms
    • C) Nitrogen (N\text{N}), and it contains 3.01×10233.01 \times 10^{23} atoms
    • D) Silicon (Si\text{Si}), and it contains 6.02×10236.02 \times 10^{23} atoms
  5. 5.

    A chemistry student prepares four closed containers of pure gases to investigate the relationship between moles, molecules, and total atoms using Avogadro's number (NA=6.02×1023 mol−1N_A = 6.02 \times 10^{23}\text{ mol}^{-1}). Container W holds 3.01×10233.01 \times 10^{23} molecules of oxygen gas (O2\text{O}_2), Container X holds 1.204×10241.204 \times 10^{24} atoms of helium gas (He\text{He}), Container Y holds 0.50 mol of carbon dioxide gas (CO2\text{CO}_2), and Container Z holds 0.25 mol of methane gas (CH4\text{CH}_4). Based on these data, which conclusion regarding the microscopic particle counts of the samples is correct?

    • A) Container W contains the greatest total number of atoms among the four samples.
    • B) Container Y and Container Z contain an identical number of gas molecules.
    • C) Container Y contains 1.5 times as many total atoms as Container W.
    • D) Container X contains fewer total fundamental particles than Container Z.

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