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CHEMISTRY, M2 EQ-Bank 12

  1. Calculate the amount (in moles) of a solute in a 2.0 L solution with a concentration of 0.75 mol/L.   (1 mark)
  1. Using your answer from part a, determine the mass of the solute if the solute is \(\ce{NaCl}\).   (2 marks)
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a.    \(1.5\ \text{mol}\)

b.    \(87.99\ \text{g}\)

Show Worked Solution

a.    \(n = c \times V = 0.75 \times 2 = 1.5\ \text{mol}\)
 

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

\(m\ce{(NaCl)} = MM \times n = 58.44 \times 1.5 = 87.99\ \text{g}\)

CHEMISTRY, M2 EQ-Bank 11

A student is investigating the concentration of copper ions in a water sample collected from a local river. They use an instrument to determine that the sample contains copper ions at a concentration level of 1.75 ppm.

  1. Calculate the mass of copper ions in a 2 L sample of water.   (2 marks)
  1. Explain why parts per million is a suitable unit for measuring low concentrations of ions in environmental samples like river water.   (2 marks)
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a.    3.5 mg

b.    Benefits of using ppm as a concentration measurement.

  • Parts per million (ppm) is suitable for measuring low concentrations of ions because it represents the mass of solute per million parts of solution, making it ideal for detecting trace levels of substances.
  • This unit allows for easy comparison of small concentrations, which is especially useful in environmental studies where contaminants are present in very low amounts.
Show Worked Solution

a.    \(1\ \text{ppm} = 1\ \text{mg/L}\)

\(\Rightarrow \ce{[Cu^{2+}]} = 1.75\ \text{mg/L}\)

\(m\ce{(Cu^{2+})} = 1.75 \times 2 = 3.5\ \text{mg}\)
 

b.    Benefits of using ppm as a concentration measurement.

  • Parts per million (ppm) is suitable for measuring low concentrations of ions because it represents the mass of solute per million parts of solution, making it ideal for detecting trace levels of substances.
  • This unit allows for easy comparison of small concentrations, which is especially useful in environmental studies where contaminants are present in very low amounts.

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

  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 \(=0.025 + 0.040 = 0.065\ \text{L}\)

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

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