Quiz: Solutions and Ideal Gases — 18 questions

Detailed questions and answers

1. Which expression gives the mass of solute needed to prepare a solution of concentration CsoluteC_{\text{solute}} and volume VsolutionV_{\text{solution}}?

msolute=CsoluteMsoluteVsolutionm_{\text{solute}}=\frac{C_{\text{solute}}M_{\text{solute}}}{V_{\text{solution}}}
msolute=Csolute+Vsolution+Msolutem_{\text{solute}}=C_{\text{solute}}+V_{\text{solution}}+M_{\text{solute}}
msolute=CsoluteVsolutionMsolutem_{\text{solute}}=C_{\text{solute}}V_{\text{solution}}M_{\text{solute}}
msolute=VsolutionCsoluteMsolutem_{\text{solute}}=\frac{V_{\text{solution}}}{C_{\text{solute}}M_{\text{solute}}}

$$m_{\text{solute}}=C_{\text{solute}}V_{\text{solution}}M_{\text{solute}}$$

Explanation

The required mass equals the amount of solute multiplied by its molar mass, with the amount given by concentration times solution volume. Dividing by volume would produce the wrong dependence because a larger final volume requires more solute at fixed concentration.

2. A student begins with a measured mass of a solid compound and dissolves it in a chosen volume of solvent; which preparation method is being used?

Preparation by dissolution
Concentration by evaporation
Preparation by dilution
Separation by filtration

Preparation by dissolution

Explanation

Dissolution prepares a solution by determining the required mass of a solid solute and dissolving it in the selected solvent volume. Dilution instead begins with a mother solution and uses it to make a less concentrated daughter solution.

3. What happens to an ionic solid when it dissolves in water?

Its solute amount changes into a different chemical element
Its solution volume becomes equal to its solute mass
Its solid structure separates into aqueous ions
Its particles remain as solid units suspended in water

Its solid structure separates into aqueous ions

Explanation

Water separates the ions that made up the ionic solid, producing aqueous ions throughout the solution. A dilution changes concentration by adding solvent but does not describe this separation into new aqueous ions.

4. Which aqueous equation correctly represents the dissolution of iron(III) chloride?

FeCl3(aq)waterFe3(aq)+3Cl+(aq)\mathrm{FeCl_3(aq)}\xrightarrow{\text{water}}\mathrm{Fe^{3-}(aq)}+3\mathrm{Cl^+(aq)}
FeCl3(s)waterFe2+(aq)+3Cl(aq)\mathrm{FeCl_3(s)}\xrightarrow{\text{water}}\mathrm{Fe^{2+}(aq)}+3\mathrm{Cl^-(aq)}
FeCl3(s)waterFe+(aq)+Cl33(aq)\mathrm{FeCl_3(s)}\xrightarrow{\text{water}}\mathrm{Fe^{+}(aq)}+\mathrm{Cl_3^{3-}(aq)}
FeCl3(s)waterFe3+(aq)+3Cl(aq)\mathrm{FeCl_3(s)}\xrightarrow{\text{water}}\mathrm{Fe^{3+}(aq)}+3\mathrm{Cl^-(aq)}

$$\mathrm{FeCl_3(s)}\xrightarrow{\text{water}}\mathrm{Fe^{3+}(aq)}+3\mathrm{Cl^-(aq)}$$

Explanation

Each formula unit of iron(III) chloride produces one Fe3+\mathrm{Fe^{3+}} ion and three Cl\mathrm{Cl^-} ions in water. The formula FeCl3\mathrm{FeCl_3} represents the solid compound, not an aqueous ion with a combined charge.

5. If an iron(III) chloride solution has concentration C(FeCl3)=0.20molL1C(\mathrm{FeCl_3})=0.20\,\mathrm{mol\,L^{-1}}, what is its chloride ion concentration?

0.067molL10.067\,\mathrm{mol\,L^{-1}}
0.40molL10.40\,\mathrm{mol\,L^{-1}}
0.20molL10.20\,\mathrm{mol\,L^{-1}}
0.60molL10.60\,\mathrm{mol\,L^{-1}}

$$0.60\,\mathrm{mol\,L^{-1}}$$

Explanation

Each dissolved FeCl3\mathrm{FeCl_3} unit produces three chloride ions, so [Cl]=3C(FeCl3)=3×0.20=0.60molL1[\mathrm{Cl^-}]=3C(\mathrm{FeCl_3})=3\times0.20=0.60\,\mathrm{mol\,L^{-1}}. The value 0.20molL10.20\,\mathrm{mol\,L^{-1}} would describe the iron(III) ion concentration instead.

6. A mother solution is diluted without losing solute. Which relationship connects the mother and daughter solutions?

CdaughterVmother=CmotherVdaughterC_{\text{daughter}}V_{\text{mother}}=C_{\text{mother}}V_{\text{daughter}}
CdaughterVdaughter=CmotherVmother\frac{C_{\text{daughter}}}{V_{\text{daughter}}}=\frac{C_{\text{mother}}}{V_{\text{mother}}}
Cdaughter+Vdaughter=Cmother+VmotherC_{\text{daughter}}+V_{\text{daughter}}=C_{\text{mother}}+V_{\text{mother}}
CdaughterVdaughter=CmotherVmotherC_{\text{daughter}}V_{\text{daughter}}=C_{\text{mother}}V_{\text{mother}}

$$C_{\text{daughter}}V_{\text{daughter}}=C_{\text{mother}}V_{\text{mother}}$$

Explanation

Dilution conserves the amount of solute, which is represented by concentration multiplied by volume on each side. The daughter concentration decreases because its volume increases while this product remains constant.

7. A dilution produces a daughter solution with volume VdaughterV_{\text{daughter}} from a mother solution volume VmotherV_{\text{mother}}; how is the dilution factor defined?

F=Cmother+Cdaughter+Vdaughter+VmotherF=C_{\text{mother}}+C_{\text{daughter}}+V_{\text{daughter}}+V_{\text{mother}}
F=VmotherVdaughter=CmotherCdaughterF=\frac{V_{\text{mother}}}{V_{\text{daughter}}}=\frac{C_{\text{mother}}}{C_{\text{daughter}}}
F=VdaughterVmother=CdaughterCmotherF=\frac{V_{\text{daughter}}}{V_{\text{mother}}}=\frac{C_{\text{daughter}}}{C_{\text{mother}}}
F=VdaughterVmother=CmotherCdaughterF=\frac{V_{\text{daughter}}}{V_{\text{mother}}}=\frac{C_{\text{mother}}}{C_{\text{daughter}}}

$$F=\frac{V_{\text{daughter}}}{V_{\text{mother}}}=\frac{C_{\text{mother}}}{C_{\text{daughter}}}$$

Explanation

The dilution factor compares the daughter and mother volumes, equivalently the mother and daughter concentrations. The concentration ratio must be inverted relative to the volume ratio because dilution lowers concentration as volume increases.

8. What does a 0.9% saline solution contain in every 100 g of solution?

0.9 g of water vapor
9.0 g of dissolved salt
0.9 g of dissolved salt
90 g of dissolved salt

0.9 g of dissolved salt

Explanation

Mass percentage specifies the mass of solute present in 100 g of solution, so the solution contains 0.9 g of salt in that basis. A concentration measure does not necessarily use 100 g of solution as its reference amount.

9. How does mass percentage differ from mass fraction for the same solution?

Mass percentage is the mass fraction divided by 100 and expressed without units.
Mass percentage is the mass fraction multiplied by 100 and expressed in percent.
Mass percentage uses solute volume, whereas mass fraction uses solution volume.
Mass percentage measures density, whereas mass fraction measures total solution mass.

Mass percentage is the mass fraction multiplied by 100 and expressed in percent.

Explanation

The mass fraction is the dimensionless ratio of solute mass to solution mass, while mass percentage multiplies that ratio by 100 and reports it as a percentage. A mass fraction is therefore not obtained by dividing the percentage by another mass-based quantity.

10. Which description best matches the ideal-gas model?

Gas particles are closely packed, with strong attractions and a fixed collective volume.
Gas particles are far apart, with negligible interactions and negligible total particle volume.
Gas particles occupy most of the container, with collisions that reduce their kinetic energy.
Gas particles have moderate spacing, with interactions that determine their boiling point.

Gas particles are far apart, with negligible interactions and negligible total particle volume.

Explanation

The ideal-gas model treats particles as sufficiently separated that their interactions and combined proper volume are negligible compared with the gas volume. Strong particle attractions and substantial particle volume are characteristics that make real gases depart from this model.

11. A gas sample has pressure PP, volume VV, amount nn, and absolute temperature TT; which equation relates these quantities for an ideal gas?

P+V=nRTP+V=nRT
PV=nRTPV=nRT
P=nVRTP=nVRT
PV=nRTPV=\frac{n}{RT}

$$PV=nRT$$

Explanation

The ideal-gas law states that pressure times volume equals amount of substance times the gas constant and absolute temperature. The other expressions do not preserve the required relationships among these variables.

12. What does the molar volume VmV_m of an ideal gas represent?

The pressure required to compress one mole of the gas
The total volume occupied by the entire gas sample
The volume occupied by one mole of the gas
The mass contained in one liter of the gas

The volume occupied by one mole of the gas

Explanation

Molar volume is defined as the volume occupied by one mole and is given for an ideal gas by Vm=RTPV_m=\frac{RT}{P}. The total sample volume depends on how many moles are present and is therefore distinct from molar volume.

13. At fixed temperature and pressure, how can the amount of gas be calculated from its volume and molar volume?

n=VVmn=\frac{V}{V_m}
n=VmVn=\frac{V_m}{V}
n=V+Vmn=V+V_m
n=VVmn=V V_m

$$n=\frac{V}{V_m}$$

Explanation

At fixed temperature and pressure, dividing the sample volume by the volume occupied per mole gives the amount of substance. Multiplying or reversing the ratio would produce the wrong dependence on sample volume and molar volume.

14. What is the molar volume of an ideal gas at 20 °C and 1013 hPa?

1013 L mol⁻¹
22.4 L mol⁻¹
20.0 L mol⁻¹
24.1 L mol⁻¹

24.1 L mol⁻¹

Explanation

At 20 °C and 1013 hPa, the molar volume is 24.1 L mol⁻¹. The value 22.4 L mol⁻¹ applies at 0 °C and 1013 hPa, so it corresponds to different conditions.

15. Why is hydrochloric acid added in excess when determining the calcium carbonate content of soil?

To increase the molar mass of the sample
To convert calcium carbonate into oxygen
To ensure that all calcium carbonate reacts
To prevent carbon dioxide from dissolving

To ensure that all calcium carbonate reacts

Explanation

Excess hydrochloric acid promotes complete reaction of the calcium carbonate, allowing the released carbon dioxide to represent the full carbonate content. Using too little acid could leave some calcium carbonate unreacted and produce an underestimated result.

16. What amount of carbon dioxide corresponds to a volume of 72 mL at 20 °C and 1013 hPa, using a molar volume of 24.1 L mol⁻¹?

7.2×102mol7.2\times10^{-2}\,\mathrm{mol}
2.4×102mol2.4\times10^{-2}\,\mathrm{mol}
3.0×101mol3.0\times10^{-1}\,\mathrm{mol}
3.0×103mol3.0\times10^{-3}\,\mathrm{mol}

$$3.0\times10^{-3}\,\mathrm{mol}$$

Explanation

Converting 72 mL to 0.072 L and dividing by 24.1 L mol⁻¹ gives nCO2=3.0×103moln_{\mathrm{CO_2}}=3.0\times10^{-3}\,\mathrm{mol}. The larger values result from failing to apply the volume conversion or from using an incorrect relationship.

17. Given nCO2=3.0×103moln_{\mathrm{CO_2}}=3.0\times10^{-3}\,\mathrm{mol} and a one-to-one reaction ratio, what mass of calcium carbonate is produced?

1.20g1.20\,\mathrm{g}
0.30g0.30\,\mathrm{g}
0.025g0.025\,\mathrm{g}
3.00g3.00\,\mathrm{g}

$$0.30\,\mathrm{g}$$

Explanation

The stoichiometry gives the calcium carbonate amount as 3.0×103mol3.0\times10^{-3}\,\mathrm{mol}, and multiplying by its molar mass gives a mass of 0.30g0.30\,\mathrm{g}. The value of 1.20g1.20\,\mathrm{g} refers to the original soil sample rather than the calcium carbonate it contains.

18. If a 1.2 g soil sample contains 0.30 g of calcium carbonate, what is the calcium carbonate mass percentage?

40%
4.0%
25%
75%

25%

Explanation

The mass percentage is calculated as 0.301.2×100=25%\frac{0.30}{1.2}\times100=25\%. A value such as 40% results from reversing or misapplying the masses in the percentage calculation.

Review with flashcards

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What formula gives the mass of solute to dissolve in a solution?

msolute=nsoluteMsolute=CsoluteVsolutionMsolutem_{\text{solute}}=n_{\text{solute}}M_{\text{solute}}=C_{\text{solute}}V_{\text{solution}}M_{\text{solute}}

What is the first step in preparing a solution by dissolution?

Determining the required mass of solid solute.

What is the second step in preparing a solution by dissolution?

Dissolving the solute in the chosen solvent volume.

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