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Avogadro's number

Chemistry • The Mole • Transitional Reasoning • 5 questions

Avogadro's number Transitional Reasoning Download PDF

Chemistry · The Mole & Stoichiometry · Chemistry · The Mole

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

Questions

  1. 1.

    A quality-control chemist prepares a standardized electrolyte solution by completely dissolving 0.050 mol0.050\text{ mol} of solid iron(III) sulfate, Fe2(SO4)3\text{Fe}_2(\text{SO}_4)_3, in deionized water. Complete dissociation yields aqueous iron(III) cations and sulfate anions according to the stoichiometry of the formula unit. Taking Avogadro's number as NA=6.02×1023 mol−1N_A = 6.02 \times 10^{23}\text{ mol}^{-1}, which pair of values represents the total number of dissolved ions and the total number of oxygen atoms originating from the solute?

    • A) 3.01×10223.01 \times 10^{22} total ions and 1.20×10231.20 \times 10^{23} oxygen atoms
    • B) 1.51×10231.51 \times 10^{23} total ions and 1.20×10231.20 \times 10^{23} oxygen atoms
    • C) 1.51×10231.51 \times 10^{23} total ions and 3.61×10233.61 \times 10^{23} oxygen atoms
    • D) 3.01×10233.01 \times 10^{23} total ions and 5.12×10235.12 \times 10^{23} oxygen atoms
  2. 2.

    An ultra-sensitive mass spectrometer detects an individual molecule of an unbranched alkane hydrocarbon with the general formula CnH2n+2\text{C}_n\text{H}_{2n+2} having an absolute mass of 7.31×10−23 g7.31 \times 10^{-23}\text{ g}. Using Avogadro's number (NA=6.02×1023 mol−1N_{\text{A}} = 6.02 \times 10^{23}\text{ mol}^{-1}) and the atomic molar masses C=12.0 g/mol\text{C} = 12.0\text{ g/mol} and H=1.0 g/mol\text{H} = 1.0\text{ g/mol}, which values correctly identify the molar mass and molecular formula of this alkane?

    • A) 16 g/mol and CH4\text{CH}_4
    • B) 30 g/mol and C2H6\text{C}_2\text{H}_6
    • C) 58 g/mol and C4H10\text{C}_4\text{H}_{10}
    • D) 44 g/mol and C3H8\text{C}_3\text{H}_8
  3. 3.

    A semiconductor laboratory deposits a high-purity monatomic metallic film onto a microchip component. A surface analysis instrument determines that the deposited layer contains exactly 3.01×10203.01 \times 10^{20} atoms of the metal and has a total mass of 0.0540 g0.0540\text{ g}. Taking Avogadro's number as NA=6.02×1023 mol−1N_A = 6.02 \times 10^{23}\text{ mol}^{-1}, what are the molar amount and the molar mass of the deposited metal?

    • A) 2.00×10−3 mol2.00 \times 10^{-3}\text{ mol} and 27.0 g/mol27.0\text{ g/mol}
    • B) 5.00×10−3 mol5.00 \times 10^{-3}\text{ mol} and 10.8 g/mol10.8\text{ g/mol}
    • C) 1.00×10−3 mol1.00 \times 10^{-3}\text{ mol} and 54.0 g/mol54.0\text{ g/mol}
    • D) 5.00×10−4 mol5.00 \times 10^{-4}\text{ mol} and 108 g/mol108\text{ g/mol}
  4. 4.

    An environmental monitoring device in a facility records a release of dinitrogen monoxide (N2O\text{N}_2\text{O}) gas that contains exactly 6.02×10226.02 \times 10^{22} nitrogen atoms. To assess the emission, the operator must determine the molar amount of the compound emitted and the total number of N2O\text{N}_2\text{O} molecules present. Taking Avogadro's number as NA=6.02×1023 mol−1N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}, which pair of values accurately represents the molar amount and the molecule count of this gas sample?

    • A) 0.10 mol0.10\ \text{mol} of N2O\text{N}_2\text{O} and 6.02×10226.02 \times 10^{22} molecules
    • B) 0.05 mol0.05\ \text{mol} of N2O\text{N}_2\text{O} and 3.01×10223.01 \times 10^{22} molecules
    • C) 0.20 mol0.20\ \text{mol} of N2O\text{N}_2\text{O} and 1.204×10231.204 \times 10^{23} molecules
    • D) 0.05 mol0.05\ \text{mol} of N2O\text{N}_2\text{O} and 9.03×10229.03 \times 10^{22} molecules
  5. 5.

    A researcher prepares two sealed containers: Container X contains 0.60 mol of argon gas (Ar), and Container Y contains 0.20 mol of sulfur trioxide gas (SO3\text{SO}_3). If Avogadro's number is represented by NAN_A, which statement correctly evaluates the particle and atom counts in the two containers?

    • A) Container X contains three times as many gas particles as Container Y, but Container Y contains a greater total number of atoms (0.80NA0.80 N_A) than Container X (0.60NA0.60 N_A).
    • B) Container X contains three times as many gas particles as Container Y, and Container X also contains a greater total number of atoms (1.80NA1.80 N_A) than Container Y (0.80NA0.80 N_A).
    • C) Both containers contain an equal total number of atoms because 0.60 mol of monoatomic argon equals the total atom count in 0.20 mol of sulfur trioxide.
    • D) Container X contains fewer gas particles than Container Y, but Container X contains a greater total number of atoms than Container Y.