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CHEMISTRY, M2 EQ-Bank 11 MC

A student prepares a solution of potassium nitrate by dissolving 0.05 kg of \(\ce{KNO3}\) in enough water to make 2000 mL of solution. Which of the following correctly calculates the concentration of the solution in mol L\(^{-1}\)?

  1. \(25 \times 10^{-6}\ \text{g L}^{-1}\)
  2. \(25 \times 10^{-4}\ \text{g L}^{-1}\)
  3. \(25 \times 10^{-2}\ \text{g L}^{-1}\)
  4. \(25\ \text{g L}^{-1}\)
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\(D\)

Show Worked Solution
  • Mass of \(\ce{KNO3}\) = 0.05 × 1000 = 50 g.
  • Volume of solution = 2 litres (L)
  •    \(c\ce{(KNO3)} = \dfrac{50}{2} = 25\ \text{g L}^{-1}\)

\(\Rightarrow D\)

CHEMISTRY, M2 EQ-Bank 10

  1. A student is asked to prepare 500.0 mL of a 0.150 mol L\(^{-1}\) standard solution of oxalic acid \(\ce{(C2H2O4.2H2O)}\), and then to perform a dilution to produce 250.0 mL of a 0.0300 mol L\(^{-1}\) solution. Outline and explain each step in this process, including the calculations involved and choice of equipment.   (5 marks)
  1. Justify the procedure in part (a.) by explaining two measures taken to ensure the accuracy of the standard solution and diluted solution produced.   (2 marks)
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a.    Calculate the mass of oxalic acid:

\(MM\ce{(C2H2O4.2H2O)}= 2(12.01) + 2(1.008) + 4(16.00) + 4(1.008) +2(16.00) = 126.068\ \text{g mol}^{-1}\)

\(n\ce{(C2H2O4.2H2O)} = c \times V = 0.150 \times 0.5 = 0.075\ \text{mol}\)

\(m\ce{(C2H2O4.2H2O)} = 0.075 \times 126.068 = 9.455\ \text{g}\)

  • Measure 9.455 g of oxalic acid using an electronic balance.

Prepare the standard solution:

  • Transfer the oxalic acid to a 500.0 mL volumetric flask using a funnel and rinse any remaining crystals from the weighing container into the flask with a small amount of distilled water.
  • Fill the volumetric flask with distilled water to about 80% full and swirl to dissolve the oxalic acid completely and carefully add more distilled water with a pipette until the bottom of the meniscus rests on the 500.0 mL mark to ensure precise volume.
  • Stopper the flask and invert several times to ensure a homogeneous solution.

Perform the dilution:

\(V_1=\dfrac{c_2V_2}{c_1} = \dfrac{0.03 \times 250}{0.15} = 50\ \text{mL}\)

  • Use a pipette to transfer 50.0 mL of the 0.150 mol L\(^{-1}\) solution into a 250.0 mL volumetric flask.
  • Dilute with distilled water up to the 250.0 mL mark in the flask to achieve a concentration of 0.0300 mol L\(^{-1}\).

b.    Answers could include two of the following:

  • A volumetric flask is used for both the standard solution and the diluted solution, as it provides precise measurements for the final solution volume. This accuracy is essential for ensuring the concentration is exactly as calculated.
  • The primary solute is weighed on an electronic balance to the nearest 0.01 g or better, minimising any error in the amount of solute added to the solution. 
  • The pipette provides precise measurements crucial for accurate dilutions.
  • Using oxalic acid as the primary standard for the investigation. Primary standard’s have high molar masses and are anhydrous (don’t absorb water). This ensures the substance is pure and the electronic balance can accurately weigh the sample.
Show Worked Solution

a.    Calculate the mass of oxalic acid:

\(MM\ce{(C2H2O4.2H2O)}= 2(12.01) + 2(1.008) + 4(16.00) + 4(1.008) +2(16.00) = 126.068\ \text{g mol}^{-1}\)

\(n\ce{(C2H2O4.2H2O)} = c \times V = 0.150 \times 0.5 = 0.075\ \text{mol}\)

\(m\ce{(C2H2O4.2H2O)} = 0.075 \times 126.068 = 9.455\ \text{g}\)

  • Measure 9.455 g of oxalic acid using an electronic balance.

Prepare the standard solution:

  • Transfer the oxalic acid to a 500.0 mL volumetric flask using a funnel and rinse any remaining crystals from the weighing container into the flask with a small amount of distilled water.
  • Fill the volumetric flask with distilled water to about 80% full and swirl to dissolve the oxalic acid completely and carefully add more distilled water with a pipette until the bottom of the meniscus rests on the 500.0 mL mark to ensure precise volume.
  • Stopper the flask and invert several times to ensure a homogeneous solution.

Perform the dilution:

\(V_1=\dfrac{c_2V_2}{c_1} = \dfrac{0.03 \times 250}{0.15} = 50\ \text{mL}\)

  • Use a pipette to transfer 50.0 mL of the 0.150 mol L\(^{-1}\) solution into a 250.0 mL volumetric flask.
  • Dilute with distilled water up to the 250.0 mL mark in the flask to achieve a concentration of 0.0300 mol L\(^{-1}\).

b.    Answers could include two of the following:

  • A volumetric flask is used for both the standard solution and the diluted solution, as it provides precise measurements for the final solution volume. This accuracy is essential for ensuring the concentration is exactly as calculated.
  • The primary solute is weighed on an electronic balance to the nearest 0.01 g or better, minimising any error in the amount of solute added to the solution. 
  • The pipette provides precise measurements crucial for accurate dilutions.
  • Using oxalic acid as the primary standard for the investigation. Primary standard’s have high molar masses and are anhydrous (don’t absorb water). This ensures the substance is pure and the electronic balance can accurately weigh the sample.

CHEMISTRY, M2 EQ-Bank 8 MC

A student prepares a standard solution of sodium chloride. They dissolve 5.85 g of sodium chloride \(\ce{(NaCl)}\) in enough water to make 1.00 L of solution. Determine the concentration of this solution.

  1. 0.100 mol L\(^{-1}\)
  2. 0.500 mol L\(^{-1}\)
  3. 1.00 mol L\(^{-1}\)
  4. 5.85 mol L\(^{-1}\)
Show Answers Only

\(A\)

Show Worked Solution

\(n\ce{(NaCl)} = \dfrac{5.85}{22.99 + 35.45} = 0.100\ \text{mol}\)

\(\ce{[NaCl]} = \dfrac{n}{V} = \dfrac{0.100}{1} = 0.100\ \text{mol L}^{-1}\)

\(\Rightarrow A\)

CHEMISTRY, M2 EQ-Bank 9

Sulfuric acid and potassium carbonate undergo neutralisation according to the following equation:

\(\ce{H2SO4(aq) + K2CO3(aq) -> K2SO4(aq) + CO2(g) + H2O(l)}\)

  1. Calculate the mass of potassium carbonate required to prepare 200 mL of a 0.15 mol L\(^{-1}\) solution of potassium carbonate.   (2 marks)
  1. 30.00 mL of the solution in part (a.) was used to neutralise 20.00 mL of sulfuric acid. Calculate the concentration of the acid.   (2 marks)
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a.    \(4.1\ \text{g}\)

b.    \(0.225\ \text{mol L}^{-1}\)

Show Worked Solution

a.    \(MM\ce{(K2CO3)} = 2(39.10) + 12.01 + 3(16.00) = 138.21\ \text{g mol}^{-1}\)

\(n\ce{(K2CO3)} = c \times V = 0.15 \times 0.2 = 0.03\ \text{mol}\)

\(\Rightarrow m\ce{(K2CO3)} = 138.21 \times 0.03 = 4.1\ \text{g}\)

 

b.    The number of moles of \(\ce{K2CO3}\) used to neutralise:

\(c \times V = 0.15 \times 0.03 = 0.0045\ \text{mol} = n\ce{(H2SO4)}\)

\(\ce{[H2SO4]} = \dfrac{0.0045}{0.02} = 0.225\ \text{mol L}^{-1}\)

CHEMISTRY, M2 EQ-Bank 7 MC

A solution is prepared by dissolving 2.00 g of calcium hydroxide \(\ce{(Ca(OH)2)}\) in enough water to produce a 500.0 mL solution. What is the concentration of hydroxide ions in the final solution?

  1. 0.108 mol L\(^{-1}\)
  2. 0.0540 mol L\(^{-1}\)
  3. 0.0270 mol L\(^{-1}\)
  4. 0.0162 mol L\(^{-1}\)
Show Answers Only

\(A\)

Show Worked Solution

\(n\ce{(Ca(OH)2)} = \dfrac{2.00}{40.08 + 2(1.008) + 2(16.00)} = 0.0270\ \text{mol}\)

\(\ce{Ca(OH)2(s) -> Ca^{2+}(aq) + 2OH-(aq)}\)

Using molar ratios:

\(n\ce{(OH^-)} = 0.0270 \times 2 = 0.0540\ \text{mol}\)

\(c\ce{(OH^-)} = \dfrac{n}{V} = \dfrac{0.0540}{0.5} = 0.108\ \text{mol L}^{-1}\)

\(\Rightarrow A\)

CHEMISTRY, M2 EQ-Bank 8

Calculate the concentration of sulfate ions present in a 750.0 mL aqueous solution containing 20.0 g of dissolved iron\(\text{(III)}\) sulfate.   (3 marks)

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\(0.200\ \text{mol L}^{-1}\)

Show Worked Solution

\(\ce{Fe2(SO4)3(s) -> 2Fe^{2+}(aq) + 3SO4^{2-}(aq)}\)

\(n\ce{(Fe2(SO4)3)} = \dfrac{m}{MM} = \dfrac{20.00}{2(55.85) + 3(32.07) + 12(16.00)} = 0.0500\ \text{mol}\)

Molar ratio: \(\ce{Fe2(SO4)3:SO4^{2-}}\ =\ 1:3\)

\(\Rightarrow n\ce{(SO4^{2-})}= 3 \times 0.0500 = 0.1500\ \text{mol}\)

\(c\ce{(SO4^{2-})} = \dfrac{0.1500}{0.75} = 0.200\ \text{mol L}^{-1}\)

CHEMISTRY, M2 EQ-Bank 7

  1. 4.56 g of potassium chloride \(\ce{KCl}\) is dissolved in 250 mL of water. What is the concentration of this solution?   (2 marks)
  1. How many grams of calcium chloride \(\ce{CaCl2}\) will be needed to make 1.50 L of a 0.250 M solution?   (2 marks)
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a.    \(0.24\ \text{mol L}^{-1}\)

b.    \(41.6\ \text{g}\)

Show Worked Solution

a.    \(MM\ce{(KCl)}= 39.10 + 35.45 = 74.55\ \text{g mol}^{-1}\)

\(n\ce{(KCl)}= \dfrac{m}{MM} = \dfrac{4.56}{74.55} = 0.061\ \text{mol}\)

\(c\ce{(KCl)}=\dfrac{n}{V}= \dfrac{0.061}{0.25} = 0.24\ \text{mol L}^{-1}\)

 

b.    \(n\ce{(CaCl2)} = c \times V = 0.25 \times 1.5 = 0.375\ \text{mol}\)

\(m\ce{(CaCl2)} = MM \times n = (40.08 + 2(35.45)) \times 0.375 = 41.6\ \text{g}\)

CHEMISTRY, M2 EQ-Bank 6

Iron \(\text{(III)}\) hydroxide can be precipitated from the reaction of iron \(\text{(III)}\) nitrate solution with sodium hydroxide solution.

\(\ce{Fe(NO3)3(aq) + 3NaOH(aq) -> Fe(OH)3(s) + 3NaNO3(aq)}\)

  1. Calculate the mass of precipitate formed when 25.0 mL of 0.150 mol L\(^{-1}\) iron \(\text{(III)}\) nitrate solution is added to 40.0 mL of 0.250 mol L\(^{-1}\) sodium hydroxide solution.   (3 marks)
  1. Calculate the concentration of nitrate ions in the final solution.   (2 marks)
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a.    \(0.356\ \text{g}\)

b.    \(0.173\ \text{mol L}^{-1}\)

Show Worked Solution

a.    \(n\ce{(Fe(NO3)3)} = c \times V = 0.150 \times 0.025 = 0.00375\ \text{mol}\)

\(n\ce{(NaOH)} = 0.25 \times 0.040 = 0.01\ \text{mol}\)

  • The molar ratio of  \(\ce{Fe(NO3)3:NaOH} = 1:3\)
  • \(3 \times 0.00375 = 0.01125\ \text{mol}\) of \(\ce{NaOH}\) is required to react with 0.00375 mol of \(\ce{Fe(NO3)3}\).
  • Due to there only being 0.01 mol of \(\ce{NaOH}\) present, \(\ce{NaOH}\) will be the limiting reagent. 
  • \(n\ce{(Fe(OH)3)}\) formed \(=n\ce{(NaOH)} \times \dfrac{1}{3} = 0.01 \times \dfrac{1}{3} = 0.00333\ \text{mol}\)
  • \(m\ce{(Fe(OH)3)}= n \times MM = 0.00333 \times (55.85 + 3(1.008) + 3(16.00))= 0.356\ \text{g}\)

 

b.    Volume of final solution \(=0.025 + 0.040 = 0.065\ \text{L}\)

\(n\ce{(NO3^-)}= 3 \times n(Fe^{3+}) = 3 \times 0.00375 = 0.01125\ \text{mol}\)

\(c\ce{(NO3^-)} = \dfrac{n}{V} = \dfrac{0.01125}{0.065} = 0.173\ \text{mol L}^{-1}\)

CHEMISTRY, M2 EQ-Bank 5

The compound potassium nitrate has the formula \(\ce{KNO3}\).

  1. A student makes a solution of this compound by dissolving 40.0 g in 250.0 mL of distilled water. Calculate the concentration of this solution in mol L\(^{-1}\).   (2 marks)
  1. The student now requires 500.0 mL of a 5.0% (w/v) solution. What volume of the solution in part (a) is required to make this?   (3 marks)
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a.    \(1.58\ \text{mol L}^{-1}\)

b.    \(0.156\ \text{L}\)

Show Worked Solution

a.    \(n\ce{(KNO3)}= \dfrac{m}{MM} = \dfrac{40.0}{39.10 + 14.01 + 3(16.00)} =0.396\ \text{mol}\)

Concentration of solution \(=\dfrac{n}{V} = \dfrac{0.396}{0.25} = 1.58\ \text{mol L}^{-1}\)

 

b.    A 5% solution requires 5 grams of solute in 100 mL of solution.

Therefore a 500 mL solution requires 25 grams of solute.

\(n\ce{(KNO3)} = \dfrac{25}{101.11} = 0.247\ \text{mol}\)

\(c\ce{(KNO3)} = \dfrac{n}{V} = \dfrac{0.247}{0.5} = 0.494\ \text{mol L}^{-1}\)

\(c_1V_1\) \(=c_2V_2\)  
\(1.58 \times V_1\) \(=0.494 \times 0.5\)  
\(V_1\) \(=\dfrac{0.247}{1.58}\)  
\(V_1\) \(=0.156\ \text{L}\)  

CHEMISTRY, M2 EQ-Bank 1

A solution is prepared by dissolving 5.00 g of sodium carbonate \(\ce{(Na2CO3)}\) in enough water to make 250.0 mL of solution.

Calculate the concentration of the sodium carbonate solution in mol L\(^{-1}\).   (2 marks)

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\(0.189\ \text{mol L}^{-1}\)

Show Worked Solution

\(\ce{MM(Na2CO3)}= 105.99\ \text{g mol}^{-1}\)

\(\ce{n(Na2CO3)}= \dfrac{5}{105.99} = 0.0472\ \text{mol}\)
 

Concentration of \(\ce{Na2CO3}\):

\(c= \dfrac{\text{n}}{\text{V}} = \dfrac{0.0472}{0.250}=0.189\ \text{mol L}^{-1}\)

CHEMISTRY, M2 EQ-Bank 1 MC

A glucose solution is prepared by dissolving pure glucose \(\ce{C6H12O6}\) in distilled water. A 50 mL vial contains 12.0 milligrams of glucose.

What is the concentration of glucose in the solution?

  1. \(1.3 \times 10^{-3}\ \text{mol L}^{-1}\)
  2. \(2.2 \times 10^{-2}\ \text{mol L}^{-1}\)
  3. \(3.9 \times 10^{-3}\ \text{mol L}^{-1}\)
  4. \(6.7 \times 10^{-2}\ \text{mol L}^{-1}\)
Show Answers Only

\(A\)

Show Worked Solution

\(\ce{MM(C6H12O6)} = (6 \times 12.01) + (12 \times 1.008) + (6 \times 16.00) = 180.156\ \text{g mol}^{-1}\)

\(\ce{n(C6H12O6)} = \dfrac{12.0 \times 10^{-3}}{180.156}=6.66 \times 10^{-5}\ \text{mol}\)

\(c=\dfrac{\text{n}}{\text{V}}=\dfrac{6.66 \times 10^{-5}}{50 \times 10^{-3}}=1.3 \times 10^{-3}\ \text{mol L}^{-1}\)

\(\Rightarrow A\)

CHEMISTRY, M2 EQ-Bank 2

A 500 mL solution contains 10.0 g of \(\ce{NaCl}\). Calculate the molarity of the \(\ce{NaCl}\) solution. Show all relevant calculations.   (3 marks)

Show Answers Only

\(\text{0.342 M}\)

Show Worked Solution

The molar mass of \(\ce{NaCl}\) is calculated as follows:

\( \ce{M(NaCl) = 22.99 + 35.45 = 58.44\ \text{g/mol}} \)

Calculate the moles of \(\ce{NaCl}\):

\( \ce{n(NaCl) =\dfrac{\text{m}}{\text{M}}= \dfrac{10.0}{58.44} = 0.171\ \text{mol}} \)

Finally, calculate the molarity of the solution using the volume in litres:

\(c = \dfrac{\text{n}}{\text{V}} = \dfrac{0.171}{0.500} = 0.342\ \text{M}\)

Therefore, the molarity of the \(\ce{NaCl}\) solution is \(0.342\ \text{M}\).

CHEMISTRY, M2 2018 VCE 2c

A solution of \(\ce{H2O2}\) is labelled ‘10 volume’ because 1.00 L of this solution produces 10.0 L of \(\ce{O2}\) measured at standard laboratory conditions (SLC) when the \(\ce{H2O2}\) in the solution is fully decomposed.

\(\ce{2H2O2 -> 2H2O + O2}\)

Calculate the concentration of \(\ce{H2O2}\) in the ‘10 volume’ solution, in grams per litre, when this solution is first prepared.  (3 marks)

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\(\ce{[H2O2] = 27.4\ \text{g L}^{-1}}\)

Show Worked Solution

\(\ce{n(O2) = \dfrac{10.0}{24.8} = 0.403\ \text{mol  (at SLC)}}\)

\(\ce{n(H2O2) = 2 \times n(O2) = 2 \times 0.403 = 0.806\ \text{mol}}\)

\(\ce{MM(H2O2) = 2 \times 1.008 + 2 \times 16 = 34.0\ \text{g}}\)

\(\ce{m(H2O2) = 0.806 \times 34.0 = 27.4\ \text{g}}\)

\(\therefore \ce{[H2O2] = 27.4\ \text{g L}^{-1}}\)

♦♦ Mean mark 38%.

CHEMISTRY, M2 2011 HSC 18* MC

A household cleaning agent contains a weak base with the formula \( \ce{NaX}\). 1.00 g of this compound was dissolved in water to give 100.0 mL of solution. A 20.0 mL sample of the solution was mixed with 0.100 mol L\(^{-1}\) hydrochloric acid, and required 24.4 mL of the acid for neutralisation.

\(\ce{NaX + HCl -> NaCl + HX}\)

What is the molar mass of the weak base?

  1. 82.0 g mol\(^{-1}\)
  2. 84.0 g mol\(^{-1}\)
  3. 122 g mol\(^{-1}\)
  4. 410 g mol\(^{-1}\)
Show Answers Only

\(A\)

Show Worked Solution

\(\ce{n(HCl)} = 0.100 \times 24.4 \times 10^{-3} = 2.44 \times 10^{-3}\ \text{mol}\)

\(\ce{n(NaX)}=2.44 \times 10^{-3}\ \text{mol in 20 mL sample.}\)

\(\ce{c(NaX)}=\dfrac{2.44 \times 10^{-3}}{20 \times 10^{-3}} =0.122\ \text{mol L}^{-1}\)

\(\ce{n(NaX)}=0.122 \times 0.1=0.0122\ \text{mol (in 100 mL sample)}\)

\(\ce{MM(NaX)}=\dfrac{1}{0.0122}=82\ \text{g mol}^{-1}\)

\(\Rightarrow A\)

CHEMISTRY, M2 2015 HSC 26c

The sodium hydroxide solution was mixed with 25.0 mL samples of 0.100 mol L\(^{-1}\) citric acid \(\ce{(C6H8O7)}\). The average volume of sodium hydroxide used was 41.50 mL.  

\(\ce{C6H8O7 + 3NaOH \rightarrow C6H5O7Na3 + 3 H2O}\)

Calculate the concentration of the sodium hydroxide solution.  (4 marks)

Show Answers Only

\(\ce{0.18 mol L^{-1} \text{(to 2 d.p.)}}\)

Show Worked Solution

\(\ce{n(C6H8O7) = c \times V = 0.100 \times 0.0250 = 0.00250 moles}\)

\(\ce{NaOH : C6H8O7}\ \ \text{reaction ratio is}\ 3:1\)

\(\ce{n(NaOH in 41.50 mL) = 3 \times 0.00250 = 0.00750 moles}\)

\[\ce{[NaOH] = \frac{n}{V} = \frac{0.00750}{0.04150} = 0.18072… = 0.18 mol L^{-1} \text{(to 2 d.p.)}}\]

Mean mark 56%.