Aussie Maths & Science Teachers: Save your time with SmarterEd
--- 2 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
a. \(1.5\ \text{mol}\)
b. \(87.99\ \text{g}\)
a. \(n = c \times V = 0.75 \times 2 = 1.5\ \text{mol}\)
b. \(MM\ce{(NaCl)} = 22.99 + 35.45 = 58.44\ \text{g/mol}\)
\(m\ce{(NaCl)} = MM \times n = 58.44 \times 1.5 = 87.99\ \text{g}\)
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.
--- 4 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
a. \(3.5\ \text{mg}\)
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.
a. \(1\ \text{ppm} = 1\ \text{mg/L}\)
\(\therefore \ce{[Cu^{2+}]} = 1.75\ \text{mg/L}\)
\(m\ce{(Cu^{2+})} = 1.75 \times 2 = 3.5\ \text{mg}\)
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.
Sulfuric acid and potassium carbonate undergo neutralisation according to the following equation:
\(\ce{H2SO4(aq) + K2CO3(aq) -> K2SO4(aq) + CO2(g) + H2O(l)}\)
--- 4 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
a. \(4.1\ \text{g}\)
b. \(0.225\ \text{mol L}^{-1}\)
a. \(MM\ce{(K2CO3)} = 2(39.10) + 12.01 + 3(16.00) = 138.21\ \text{g mol}^{-1}\)
\(n\ce{(K2CO3)} = c \times V = 0.15 \times 0.2 = 0.03\ \text{mol}\)
\(\therefore m\ce{(K2CO3)} = 138.21 \times 0.03 = 4.1\ \text{g}\)
b. The number of moles of \(\ce{K2CO3}\) used to neutralise:
\(c \times V = 0.15 \times 0.03 = 0.0045\ \text{mol} = n\ce{(H2SO4)}\)
\(\ce{[H2SO4]} = \dfrac{0.0045}{0.02} = 0.225\ \text{mol L}^{-1}\)
--- 4 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
a. \(0.24\ \text{mol L}^{-1}\)
b. \(41.6\ \text{g}\)
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. \(n\ce{(CaCl2)} = c \times V = 0.25 \times 1.5 = 0.375\ \text{mol}\)
\(m\ce{(CaCl2)} = MM \times n = (40.08 + 2(35.45)) \times 0.375 = 41.6\ \text{g}\)
Iron \(\text{(III)}\) hydroxide can be precipitated from the reaction of iron \(\text{(III)}\) nitrate solution with sodium hydroxide solution.
\(\ce{Fe(NO3)3(aq) + 3NaOH(aq) -> Fe(OH)3(s) + 3NaNO3(aq)}\)
--- 6 WORK AREA LINES (style=lined) ---
--- 4 WORK AREA LINES (style=lined) ---
a. \(0.356\ \text{g}\)
b. \(0.173\ \text{mol L}^{-1}\)
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 \(=0.025 + 0.040 = 0.065\ \text{L}\)
\(n\ce{(NO3^-)}= 3 \times n(Fe^{3+}) = 3 \times 0.00375 = 0.01125\ \text{mol}\)
\(c\ce{(NO3^-)} = \dfrac{n}{V} = \dfrac{0.01125}{0.065} = 0.173\ \text{mol L}^{-1}\)
A 5.0 L sample of water (density = 1.0 kg L\(^{-1}\)) is contaminated with 15 ppm of mercury ions. What is the mass of mercury in this sample?
\(A\)
→ The mass of the water is 5 kg.
→ 15 ppm = 15 mg/kg.
→ Mass of mercury in the sample \(=15 \times 5 = 75\ \text{mg}\).
\(\Rightarrow A\)