M2 Quantitative Chemistry: Original 70-Mark Revision Exam

This original tutorgum practice paper covers mixed revision from the legacy Chemistry 2017 M2: Introduction to Quantitative Chemistry unit. It tests quantitative problem-solving, chemical relationships, experimental technique, and the reasoning needed to explain your results. It is a focused topic assessment, not a full-course HSC Chemistry exam or an official, endorsed, past, or predicted NESA paper.

01Exam instructions

Total marks: 70

Suggested working time: 105 minutes

Suggested reading time: 5 minutes

Permitted equipment: Scientific calculator, pens, and pencils. No notes are required.

Assessed scope

This paper assesses the supplied legacy Chemistry 2017 quantitative chemistry content, including:

• Balancing equations, mole ratios, and conservation of mass.
• The mole, Avogadro’s constant, molar mass, and empirical formulae.
• Stoichiometric calculations, limiting reagents, and excess reactants.
• Gas laws, ideal gas calculations, and gaseous reaction volumes.
• Amount concentration, mass concentration, percentage concentration, and dilution.
• Preparing standard solutions and evaluating volumetric technique.
• Experimental mass changes and whole-number mole ratios.

It excludes other syllabus editions and topics outside this specified unit. It should not be used as a full-course HSC Chemistry assessment.

Data and assumptions

Use the following relative atomic masses, \(A_r\), where required.

Additional data:

• Avogadro constant: \(N_A = 6.02 \times 10^{23}\ \mathrm{mol^{-1}}\).
• Ideal gas constant: \(R = 8.31\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}\).
• \(1\ \mathrm{L} = 1000\ \mathrm{mL}\).
• \(n\) represents amount in moles, \(m\) represents mass, \(M\) represents molar mass, \(c\) represents amount concentration, \(V\) represents volume, \(P\) represents pressure, and \(T\) represents absolute temperature.
• Use kelvin for gas-law calculations and litres when calculating gas volumes with the ideal gas equation.
• Assume ideal gas behaviour where required. Gas-volume ratios in reactions apply when all gas volumes are measured at the same temperature and pressure.
• For percentage mass/volume concentration, \(x\%\ \mathrm{m/v}\) means \(x\) grams of solute per 100 mL of final solution.
• Unless otherwise stated, report calculated values to three significant figures and include appropriate units.

Show your working for calculation questions. A correct numerical answer without working may not earn all available marks.

02Questions

Question 1 (6 marks)

(a) Balance the following chemical equation using the smallest possible whole-number coefficients. (2 marks)

\[ \ce{Fe + O2 -> Fe2O3} \]

(b) Using your balanced equation from part (a), state the mole ratio of iron to oxygen gas to iron(III) oxide. (1 mark)

(c) The total mass of reactants in a sealed vessel is 12.00 g before a reaction. After the reaction, the total mass of the products is measured as 12.00 g.

State the relevant law and explain why the measured mass remains unchanged. (1 mark)

(d) A piece of magnesium has a mass of 0.48 g before burning. After burning, the mass of the solid product is 0.80 g. Assume all the magnesium oxide remains in the container and no other solid is added.

Calculate the mass of oxygen that combined with the magnesium, and explain why the measured mass of the solid increased. (2 marks)

Question 2 (7 marks)

Calcium hydroxide has the formula \(\ce{Ca(OH)2}\). Use the relative atomic masses supplied in the data table.

(a) Calculate the molar mass of calcium hydroxide. (2 marks)

(b) Calculate the amount, in moles, present in a 5.55 g sample of calcium hydroxide. (2 marks)

(c) Calculate the number of formula units in this sample. (1 mark)

(d) Calculate the percentage by mass of oxygen in calcium hydroxide. (2 marks)

Question 3 (6 marks)

A pure compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.

Use \(A_r(\ce{C}) = 12.0\), \(A_r(\ce{H}) = 1.0\), and \(A_r(\ce{O}) = 16.0\).

(a) Assume you have a 100 g sample. Calculate the amount, in moles, of each element in the sample. (3 marks)

(b) Determine the simplest whole-number mole ratio of the elements and hence the empirical formula of the compound. (2 marks)

(c) A separate analysis finds 0.120 mol of carbon atoms and 0.240 mol of oxygen atoms in a sample. State the simplest whole-number mole ratio of carbon to oxygen. (1 mark)

Question 4 (10 marks)

(a) A fixed amount of gas occupies 2.40 L at 100 kPa and 300 K. The gas is changed to conditions of 150 kPa and 360 K. Calculate its final volume. (2 marks)

(b) At constant pressure, the absolute temperature of a fixed amount of gas doubles. State what happens to its volume. (1 mark)

(c) A fixed amount of gas occupies a rigid container. Its pressure is 95.0 kPa at 290 K. Calculate its pressure when the temperature increases to 320 K. (2 marks)

(d) A sample contains 0.250 mol of an ideal gas at 298 K and 100 kPa. Calculate the volume of the gas using the ideal gas equation. (2 marks)

(e) Nitrogen gas reacts with hydrogen gas according to the balanced equation:

\[ \ce{N2(g) + 3H2(g) -> 2NH3(g)} \]

A volume of 12.0 L of hydrogen reacts completely with excess nitrogen. All gas volumes are measured at the same temperature and pressure.

Calculate:

(i) the volume of nitrogen consumed (1 mark)

(ii) the volume of ammonia produced (1 mark)

(iii) Explain why the balanced equation can be used to relate these gas volumes. (1 mark)

Question 5 (10 marks)

Aluminium reacts with chlorine gas to produce aluminium chloride.

\[ \ce{2Al + 3Cl2 -> 2AlCl3} \]

A reaction mixture contains 5.40 g of aluminium and 14.2 g of chlorine gas. Assume the reaction proceeds to completion, with no product loss.

Use the supplied relative atomic masses.

(a) Calculate the initial amount, in moles, of each reactant. (2 marks)

(b) Identify the limiting reagent. Justify your answer using the stoichiometric ratio of the balanced equation. (3 marks)

(c) Calculate the theoretical mass of aluminium chloride that can be produced. (3 marks)

(d) Calculate the mass of aluminium remaining after the reaction is complete. (2 marks)

Question 6 (8 marks)

(a) A sodium chloride solution contains 0.150 mol of dissolved \(\ce{NaCl}\) in a final solution volume of 500.0 mL.

Calculate its amount concentration in \(\mathrm{mol\,L^{-1}}\). (2 marks)

(b) Calculate the volume of a \(0.250\ \mathrm{mol\,L^{-1}}\) solution that contains 0.0300 mol of dissolved solute. Give your answer in millilitres. (2 marks)

(c) A solution is prepared by dissolving 8.00 g of glucose to make 200.0 mL of final solution.

Calculate its mass concentration in \(\mathrm{g\,L^{-1}}\). (1 mark)

(d) A sodium chloride solution has an amount concentration of \(0.100\ \mathrm{mol\,L^{-1}}\). The molar mass of \(\ce{NaCl}\) is \(58.5\ \mathrm{g\,mol^{-1}}\).

Calculate its mass concentration and determine whether it is equivalent to a label stating \(5.85\ \mathrm{g\,L^{-1}}\). (2 marks)

(e) A solution has a concentration of 2.0% m/v. Calculate the mass of solute in 250 mL of this solution. (1 mark)

Question 7 (6 marks)

(a) A student transfers 25.0 mL of a \(1.20\ \mathrm{mol\,L^{-1}}\) stock solution into a volumetric flask and dilutes it to a final volume of 250.0 mL.

Calculate the final amount concentration. (2 marks)

(b) What volume of a \(2.00\ \mathrm{mol\,L^{-1}}\) stock solution is required to prepare 100.0 mL of a \(0.150\ \mathrm{mol\,L^{-1}}\) solution? (2 marks)

(c) Explain why the amount of dissolved solute remains unchanged during an ideal dilution, and state the equation that relates the initial and final concentrations and volumes. (2 marks)

Question 8 (8 marks)

A student must prepare 250.0 mL of a \(0.200\ \mathrm{mol\,L^{-1}}\) standard solution of potassium nitrate, \(\ce{KNO3}\), from solid potassium nitrate.

Use the supplied relative atomic masses.

(a) Calculate the mass of potassium nitrate required. (3 marks)

(b) Outline a method for preparing the standard solution from the solid. Your method must include dissolving the solute, quantitative transfer, and adjustment to the final volume. (2 marks)

(c) The student must also dilute a stock solution by transferring 25.00 mL into a 250.0 mL volumetric flask.

Describe how to carry out this dilution accurately, including the correct use of the pipette, adjustment to the final volume, and mixing. A student has rinsed the pipette with deionised water but has not conditioned it with the stock solution.

Explain the likely effect of this error on the final concentration. (3 marks)

Question 9 (9 marks)

Calcium carbonate reacts with excess hydrochloric acid according to the balanced equation:

\[ \ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g)} \]

A student reacts 5.00 g of calcium carbonate in an open flask. The total mass of the flask contents before the reaction is 105.00 g. After the reaction is complete, the total mass of the contents is 102.80 g.

Assume that all the mass lost from the flask contents is due to carbon dioxide escaping, and no liquid is splashed out.

Use \(M(\ce{CO2}) = 44.0\ \mathrm{g\,mol^{-1}}\). Calculate any other required molar masses from the supplied relative atomic masses.

(a) Calculate the initial amount, in moles, of calcium carbonate. (2 marks)

(b) Calculate the theoretical amount and mass of carbon dioxide produced. (2 marks)

(c) Calculate the measured mass loss and compare it with the theoretical mass of carbon dioxide. (2 marks)

(d) Explain why the mass of the flask contents decreases even though the law of conservation of mass still applies. (2 marks)

(e) In a separate trial using the same initial mass of calcium carbonate, only 0.0400 mol of carbon dioxide is collected. Suggest one plausible experimental reason for this result. (1 mark)

03Worked solutions and marking guide

Award marks for the demonstrated steps and ideas listed below. Mathematically equivalent methods are acceptable. Follow-through marks may be awarded when a student applies a correct method to an earlier numerical result that was incorrect.

Solution 1

(a) Balanced equation

\[ \ce{4Fe + 3O2 -> 2Fe2O3} \]

There are four iron atoms and six oxygen atoms on each side.

Marking (2 marks):

• 1 mark for balancing iron atoms.
• 1 mark for balancing oxygen atoms.

(b) Mole ratio

\[ \ce{Fe}:\ce{O2}:\ce{Fe2O3}=4:3:2 \]

The coefficients in a balanced equation give the mole ratios of the substances reacting and forming.

Marking (1 mark):

• 1 mark for the correct ratio.

(c) Conservation of mass

The law of conservation of mass states that mass is conserved during a chemical reaction. Atoms are rearranged rather than created or destroyed. Since the vessel is sealed, matter cannot enter or leave, so the total mass remains 12.00 g.

Marking (1 mark):

• 1 mark for explaining conservation of mass in the closed system.

(d) Mass of oxygen

The increase in solid mass is the mass of oxygen incorporated into the product.

\[ \begin{aligned} m(\text{oxygen}) &= 0.80-0.48\\ &=0.32\ \mathrm{g} \end{aligned} \]

The solid gains mass because oxygen from the air combines with magnesium to form magnesium oxide.

Marking (2 marks):

• 1 mark for the mass difference of 0.32 g.
• 1 mark for explaining that oxygen from the air becomes part of the solid.

Solution 2

(a) Molar mass of calcium hydroxide

One formula unit contains one calcium atom, two oxygen atoms, and two hydrogen atoms.

\[ \begin{aligned} M(\ce{Ca(OH)2})&=40.1+2(16.0+1.0)\\ &=74.1\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Marking (2 marks):

• 1 mark for the correct sum of atomic masses.
• 1 mark for \(74.1\ \mathrm{g\,mol^{-1}}\).

(b) Amount in moles

The relationship between mass, amount, and molar mass is \(n=m/M\).

\[ \begin{aligned} n&=\frac{m}{M}\\ &=\frac{5.55}{74.1}\\ &=0.0749\ \mathrm{mol} \end{aligned} \]

Marking (2 marks):

• 1 mark for using \(n=m/M\).
• 1 mark for the correct answer with units.

(c) Number of formula units

Multiply the amount in moles by Avogadro’s constant.

\[ \begin{aligned} N&=nN_A\\ &=0.0749\times6.02\times10^{23}\\ &=4.51\times10^{22} \end{aligned} \]

The sample contains \(4.51\times10^{22}\) formula units.

Marking (1 mark):

• 1 mark for the correct number of formula units.

(d) Percentage by mass of oxygen

The total oxygen mass in one mole of calcium hydroxide is:

\[ m(\text{oxygen per mole})=2(16.0)=32.0\ \mathrm{g} \]

The percentage by mass is:

\[ \begin{aligned} \%\text{ oxygen}&=\frac{32.0}{74.1}\times100\\ &=43.2\% \end{aligned} \]

Marking (2 marks):

• 1 mark for determining the oxygen mass per mole.
• 1 mark for the correct percentage.

Solution 3

(a) Amount of each element

Assume a 100 g sample. Each percentage then corresponds to the mass of that element in grams.

For carbon:

\[ n(\ce{C})=\frac{40.0}{12.0}=3.33\ \mathrm{mol} \]

For hydrogen:

\[ n(\ce{H})=\frac{6.7}{1.0}=6.7\ \mathrm{mol} \]

For oxygen:

\[ n(\ce{O})=\frac{53.3}{16.0}=3.33\ \mathrm{mol} \]

Marking (3 marks):

• 1 mark for each correct amount.

(b) Empirical formula

Divide each amount by the smallest amount, approximately 3.33 mol.

\[ \begin{aligned} \ce{C}:&\quad 3.33/3.33\approx1\\ \ce{H}:&\quad 6.7/3.33\approx2\\ \ce{O}:&\quad 3.33/3.33\approx1 \end{aligned} \]

The simplest ratio is approximately \(1:2:1\).

Therefore, the empirical formula is \(\ce{CH2O}\).

The small departure from an exact ratio of \(1:2:1\) comes from rounding in the supplied percentages.

Marking (2 marks):

• 1 mark for the correct simplest whole-number ratio.
• 1 mark for \(\ce{CH2O}\).

(c) Carbon-to-oxygen ratio

Divide both amounts by 0.120 mol:

\[ 0.120:0.240=1:2 \]

Marking (1 mark):

• 1 mark for the ratio \(1:2\).

Solution 4

(a) Combined gas-law calculation

The amount of gas is constant. For an ideal gas undergoing a change in pressure, volume, and temperature:

\[ \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \]

Rearrange to find the final volume.

\[ \begin{aligned} V_2&=\frac{P_1V_1T_2}{P_2T_1}\\ &=\frac{100(2.40)(360)}{150(300)}\\ &=1.92\ \mathrm{L} \end{aligned} \]

Marking (2 marks):

• 1 mark for the correct relationship and substitution.
• 1 mark for \(1.92\ \mathrm{L}\).

(b) Volume and absolute temperature

At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature.

If the temperature doubles in kelvin, the volume doubles.

Marking (1 mark):

• 1 mark for stating that the volume doubles.

(c) Pressure and temperature at constant volume

At constant volume, pressure is directly proportional to absolute temperature.

\[ \begin{aligned} \frac{P_1}{T_1}&=\frac{P_2}{T_2}\\ P_2&=\frac{95.0\times320}{290}\\ &=105\ \mathrm{kPa} \end{aligned} \]

Marking (2 marks):

• 1 mark for the correct pressure-temperature relationship.
• 1 mark for the correct final pressure with units.

(d) Ideal gas equation

The ideal gas equation is \(PV=nRT\).

Rearrange to make volume the subject:

\[ \begin{aligned} V&=\frac{nRT}{P}\\ &=\frac{0.250(8.31)(298)}{100}\\ &=6.19\ \mathrm{L} \end{aligned} \]

Marking (2 marks):

• 1 mark for the correct rearrangement and substitution.
• 1 mark for \(6.19\ \mathrm{L}\).

(e) Gas volumes in a chemical reaction

The balanced equation is:

\[ \ce{N2 + 3H2 -> 2NH3} \]

At the same temperature and pressure, gaseous volumes are proportional to the amounts of gas in moles.

Three volumes of hydrogen react with one volume of nitrogen to produce two volumes of ammonia.

For nitrogen:

\[ V(\ce{N2})=\frac{12.0}{3}=4.00\ \mathrm{L} \]

For ammonia:

\[ V(\ce{NH3})=\frac{2}{3}\times12.0=8.00\ \mathrm{L} \]

The volume ratio follows the coefficient ratio because equal volumes of ideal gases at the same temperature and pressure contain equal amounts of gas. This is a consequence of Avogadro’s law.

Marking (3 marks):

• 1 mark for \(4.00\ \mathrm{L}\) of nitrogen.
• 1 mark for \(8.00\ \mathrm{L}\) of ammonia.
• 1 mark for explaining the relationship between gas volumes and mole ratios.

Solution 5

(a) Initial amounts of reactants

For aluminium:

\[ \begin{aligned} n(\ce{Al})&=\frac{5.40}{27.0}\\ &=0.200\ \mathrm{mol} \end{aligned} \]

For chlorine gas, remember that each molecule contains two chlorine atoms.

\[ M(\ce{Cl2})=2(35.5)=71.0\ \mathrm{g\,mol^{-1}} \]

\[ \begin{aligned} n(\ce{Cl2})&=\frac{14.2}{71.0}\\ &=0.200\ \mathrm{mol} \end{aligned} \]

Marking (2 marks):

• 1 mark for the amount of aluminium.
• 1 mark for the amount of chlorine gas.

(b) Limiting reagent

The balanced equation requires two moles of aluminium for every three moles of chlorine gas.

For 0.200 mol of aluminium, the required amount of chlorine gas is:

\[ n(\ce{Cl2})=0.200\times\frac{3}{2}=0.300\ \mathrm{mol} \]

Only 0.200 mol of chlorine gas is available. There is not enough chlorine to react with all the aluminium.

Therefore, chlorine gas is the limiting reagent.

Marking (3 marks):

• 1 mark for calculating the required 0.300 mol of chlorine gas.
• 1 mark for comparing the required amount with the 0.200 mol available.
• 1 mark for identifying chlorine gas as the limiting reagent.

(c) Theoretical mass of aluminium chloride

The balanced equation shows that three moles of chlorine gas produce two moles of aluminium chloride.

\[ \begin{aligned} n(\ce{AlCl3})&=0.200\times\frac{2}{3}\\ &=0.133\ \mathrm{mol} \end{aligned} \]

Calculate the molar mass of aluminium chloride:

\[ \begin{aligned} M(\ce{AlCl3})&=27.0+3(35.5)\\ &=133.5\ \mathrm{g\,mol^{-1}} \end{aligned} \]

The theoretical product mass is:

\[ \begin{aligned} m&=nM\\ &=0.133333\ldots\times133.5\\ &=17.8\ \mathrm{g} \end{aligned} \]

Marking (3 marks):

• 1 mark for the correct amount of aluminium chloride.
• 1 mark for its molar mass.
• 1 mark for the theoretical mass of 17.8 g.

(d) Excess aluminium remaining

The reaction consumes two moles of aluminium for every three moles of chlorine gas.

\[ \begin{aligned} n(\ce{Al\ consumed})&=0.200\times\frac{2}{3}\\ &=0.133333\ldots\ \mathrm{mol} \end{aligned} \]

Subtract the amount consumed from the initial amount:

\[ \begin{aligned} n(\ce{Al\ remaining})&=0.200-0.133333\ldots\\ &=0.0667\ \mathrm{mol} \end{aligned} \]

Convert the remaining amount to mass:

\[ \begin{aligned} m&=0.066666\ldots\times27.0\\ &=1.80\ \mathrm{g} \end{aligned} \]

Marking (2 marks):

• 1 mark for the amount of aluminium remaining.
• 1 mark for the remaining mass of 1.80 g.

Solution 6

(a) Amount concentration

Convert the solution volume to litres:

\[ 500.0\ \mathrm{mL}=0.5000\ \mathrm{L} \]

Amount concentration is calculated using \(c=n/V\).

\[ \begin{aligned} c&=\frac{0.150}{0.5000}\\ &=0.300\ \mathrm{mol\,L^{-1}} \end{aligned} \]

Marking (2 marks):

• 1 mark for converting the volume to litres.
• 1 mark for the correct concentration.

(b) Volume of solution

Rearrange \(c=n/V\) to find the volume:

\[ \begin{aligned} V&=\frac{n}{c}\\ &=\frac{0.0300}{0.250}\\ &=0.120\ \mathrm{L}\\ &=120\ \mathrm{mL} \end{aligned} \]

Marking (2 marks):

• 1 mark for the correct rearrangement.
• 1 mark for the correct volume with units.

(c) Mass concentration

Mass concentration is the mass of dissolved solute divided by the final volume of solution.

\[ \begin{aligned} c_m&=\frac{8.00}{0.2000}\\ &=40.0\ \mathrm{g\,L^{-1}} \end{aligned} \]

Marking (1 mark):

• 1 mark for \(40.0\ \mathrm{g\,L^{-1}}\).

(d) Comparing concentration descriptions

Convert amount concentration to mass concentration by multiplying by molar mass.

\[ \begin{aligned} c_m&=cM\\ &=0.100\times58.5\\ &=5.85\ \mathrm{g\,L^{-1}} \end{aligned} \]

The calculated value matches the label. The two descriptions are equivalent: one reports the amount of solute per litre, and the other reports its mass per litre.

Marking (2 marks):

• 1 mark for the correct conversion to \(5.85\ \mathrm{g\,L^{-1}}\).
• 1 mark for correctly comparing it with the label.

(e) Percentage mass/volume concentration

A concentration of 2.0% m/v means 2.0 g of solute per 100 mL of final solution.

\[ \begin{aligned} m&=\frac{2.0}{100}\times250\\ &=5.0\ \mathrm{g} \end{aligned} \]

Marking (1 mark):

• 1 mark for 5.0 g.

Solution 7

(a) Concentration after dilution

During dilution, the amount of solute remains constant, so:

\[ c_1V_1=c_2V_2 \]

Rearrange for the final concentration:

\[ \begin{aligned} c_2&=\frac{c_1V_1}{V_2}\\ &=\frac{1.20\times25.0}{250.0}\\ &=0.120\ \mathrm{mol\,L^{-1}} \end{aligned} \]

The two volumes can be used in millilitres because their units cancel.

Marking (2 marks):

• 1 mark for the correct dilution relationship.
• 1 mark for \(0.120\ \mathrm{mol\,L^{-1}}\).

(b) Volume of stock solution

Use the same dilution relationship and rearrange to find the initial volume.

\[ \begin{aligned} V_1&=\frac{c_2V_2}{c_1}\\ &=\frac{0.150\times100.0}{2.00}\\ &=7.50\ \mathrm{mL} \end{aligned} \]

Marking (2 marks):

• 1 mark for the correct rearrangement and substitution.
• 1 mark for 7.50 mL.

(c) Why dilution conserves solute

Adding solvent increases the total solution volume but does not change the amount of solute, provided none is lost or added.

Since \(n=cV\), the initial and final amounts are equal:

\[ n_1=n_2 \]

Therefore:

\[ c_1V_1=c_2V_2 \]

Marking (2 marks):

• 1 mark for explaining that the amount of solute remains constant.
• 1 mark for the correct equation.

Solution 8

(a) Mass of potassium nitrate

First, calculate the required amount of solute. Convert the final volume to litres.

\[ \begin{aligned} n&=cV\\ &=0.200\times0.2500\\ &=0.0500\ \mathrm{mol} \end{aligned} \]

Calculate the molar mass of potassium nitrate:

\[ \begin{aligned} M(\ce{KNO3})&=39.1+14.0+3(16.0)\\ &=101.1\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Now calculate the required mass.

\[ \begin{aligned} m&=nM\\ &=0.0500\times101.1\\ &=5.055\ \mathrm{g}\\ &\approx5.06\ \mathrm{g} \end{aligned} \]

Marking (3 marks):

• 1 mark for the amount of solute.
• 1 mark for the correct molar mass.
• 1 mark for the required mass of 5.06 g.

(b) Preparing the standard solution

A suitable method is:

1. Accurately weigh 5.06 g of potassium nitrate. Dissolve it in a small volume of deionised water in a beaker.
2. Transfer the solution quantitatively into a 250.0 mL volumetric flask. Rinse the beaker and funnel with deionised water, adding the rinsings to the flask.
3. Add deionised water until the bottom of the meniscus reaches the calibration mark at eye level. Stopper the flask and invert it several times to mix thoroughly.

The final volume must be 250.0 mL. Adding 250.0 mL of water to the solid would not produce the required final volume.

Marking (2 marks):

• 1 mark for dissolving the solid and transferring all the solute, including rinsings.
• 1 mark for making the solution to the calibration mark and mixing thoroughly.

(c) Accurate dilution and pipette error

For an accurate dilution:

1. Condition the volumetric pipette with a small portion of the stock solution after its water rinse. Use a pipette filler to measure and transfer exactly 25.00 mL into the volumetric flask.
2. Add deionised water to near the calibration mark, then add it dropwise until the bottom of the meniscus reaches the mark at eye level. Stopper the flask and invert it several times to mix.

If the pipette contains residual deionised water, that water dilutes the stock solution inside the pipette. The measured aliquot then delivers less solute than intended, so the final diluted solution will be less concentrated than the target value.

Marking (3 marks):

• 1 mark for correctly conditioning and using the volumetric pipette.
• 1 mark for accurate adjustment to the final volume and thorough mixing.
• 1 mark for identifying that residual water causes a lower-than-intended concentration.

Solution 9

(a) Initial amount of calcium carbonate

Calculate its molar mass:

\[ \begin{aligned} M(\ce{CaCO3})&=40.1+12.0+3(16.0)\\ &=100.1\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Calculate the amount in moles:

\[ \begin{aligned} n(\ce{CaCO3})&=\frac{5.00}{100.1}\\ &=0.04995\ \mathrm{mol}\\ &\approx0.0500\ \mathrm{mol} \end{aligned} \]

Marking (2 marks):

• 1 mark for the correct molar mass.
• 1 mark for the amount of calcium carbonate.

(b) Theoretical amount and mass of carbon dioxide

The balanced equation shows a mole ratio of 1:1 between calcium carbonate and carbon dioxide.

Therefore:

\[ n(\ce{CO2})=n(\ce{CaCO3})=0.04995\ \mathrm{mol} \]

Calculate the theoretical mass of carbon dioxide:

\[ \begin{aligned} m(\ce{CO2})&=nM\\ &=0.04995\times44.0\\ &=2.20\ \mathrm{g} \end{aligned} \]

Marking (2 marks):

• 1 mark for the correct mole ratio and amount of carbon dioxide.
• 1 mark for the theoretical mass of 2.20 g.

(c) Measured mass loss

The measured mass loss is:

\[ \begin{aligned} \Delta m&=105.00-102.80\\ &=2.20\ \mathrm{g} \end{aligned} \]

This matches the theoretical mass of carbon dioxide to the stated precision. The measurements are consistent with the assumption that the lost mass is carbon dioxide.

Marking (2 marks):

• 1 mark for the measured mass loss of 2.20 g.
• 1 mark for comparing it correctly with the theoretical value.

(d) Conservation of mass in an open system

The measured mass of the flask contents decreases because carbon dioxide gas escapes into the surroundings. The balance no longer measures that escaped gas as part of the flask contents.

The law of conservation of mass still applies to the complete system. If all reactants and products, including the escaped carbon dioxide, were included in a closed system, the total mass would remain constant.

Marking (2 marks):

• 1 mark for explaining that carbon dioxide leaves the measured contents.
• 1 mark for explaining that total mass is conserved when all matter is included.

(e) Lower collected amount of carbon dioxide

One plausible explanation is that some carbon dioxide escaped through a leak before it could be collected. Incomplete reaction or carbon dioxide dissolving in the liquid could also reduce the collected amount.

Marking (1 mark):

• 1 mark for one plausible experimental reason.