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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\(^{-1}\)?
\(D\)
→ Mass of \(\ce{KNO3}\) is 50 g.
→ Volume of solution is 2 L.
\(c\ce{(KNO3)} = \dfrac{50}{2} = 25\ \text{g L}^{-1}\)
\(\Rightarrow D\)
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a. Calculate the mass of oxalic acid:
\(MM\ce{(C2H2O4.2H2O)}= 2(12.01) + 2(1.008) + 4(16.00) + 4(1.008) +2(16.00) = 126.068\ \text{g mol}^{-1}\)
\(n\ce{(C2H2O4.2H2O)} = c \times V = 0.150 \times 0.5 = 0.075\ \text{mol}\)
\(m\ce{(C2H2O4.2H2O)} = 0.075 \times 126.068 = 9.455\ \text{g}\)
→ 9.455 g of oxalic acid needs to be weighed using an electronic balance.
→ 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. Measures taken to ensure an accurate and precise solution:
→ 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, minimizing 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.
a. Calculate the mass of oxalic acid:
\(MM\ce{(C2H2O4.2H2O)}= 2(12.01) + 2(1.008) + 4(16.00) + 4(1.008) +2(16.00) = 126.068\ \text{g mol}^{-1}\)
\(n\ce{(C2H2O4.2H2O)} = c \times V = 0.150 \times 0.5 = 0.075\ \text{mol}\)
\(m\ce{(C2H2O4.2H2O)} = 0.075 \times 126.068 = 9.455\ \text{g}\)
→ 9.455 g of oxalic acid needs to be weighed using an electronic balance.
→ 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. Measures taken to ensure an accurate and precise solution:
→ 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, minimizing 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.
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?
\(A\)
\(n\ce{(NaCl)} = \dfrac{5.85}{22.99 + 35.45} = 0.100\ \text{mol}\)
\(\ce{[NaCl]} = \dfrac{n}{V} = \dfrac{0.100}{1} = 0.100\ \text{mol L}^{-1}\)
\(\Rightarrow A\)
Sulfuric acid and potassium carbonate undergo neutralisation according to the following equation:
\(\ce{H2SO4(aq) + K2CO3(aq) -> K2SO4(aq) + CO2(g) + H2O(l)}\)
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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:
\(c \times V = 0.15 \times 0.03 = 0.0045\ \text{mol} = n\ce{(H2SO4)}\)
\(\ce{[H2SO4]} = \dfrac{0.0045}{0.02} = 0.225\ \text{mol L}^{-1}\)
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?
\(A\)
\(n\ce{(Ca(OH)2)} = \dfrac{2.00}{40.08 + 2(1.008) + 2(16.00)} = 0.0270\ \text{mol}\)
\(\ce{Ca(OH)2(s) -> Ca^{2+}(aq) + 2OH-(aq)}\)
Using molar ratios, \(n\ce{(OH^-)} = 0.0270 \times 2 = 0.0540\ \text{mol}\)
\(c\ce{(OH^-)} = \dfrac{n}{V} = \dfrac{0.0540}{0.5} = 0.108\ \text{mol L}^{-1}\)
\(\Rightarrow A\)
Calculate the concentration of sulfate ions present in a 750.0 mL aqueous solution containing 20.0 g of dissolved iron (III) sulfate. (3 marks)
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\(0.200\ \text{mol L}^{-1}\)
\(\ce{Fe2(SO4)3(s) -> 2Fe^{2+}(aq) + 3SO4^{2-}(aq)}\)
\(n\ce{(Fe2(SO4)3)} = \dfrac{m}{MM} = \dfrac{20.00}{2(55.85) + 3(32.07) + 12(16.00)} = 0.0500\ \text{mol}\)
The molar ratio of \(\ce{Fe2(SO4)3:SO4^{2-}}\) is \(1:3\)
\(\therefore n\ce{(SO4^{2-})}= 3 \times 0.0500 = 0.1500\ \text{mol}\)
\(c\ce{(SO4^{2-})} = \dfrac{0.1500}{0.75} = 0.200\ \text{mol L}^{-1}\)
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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.
\(\ce{Fe(NO3)3(aq) + 3NaOH(aq) -> Fe(OH)3(s) + 3NaNO3(aq)}\)
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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 \(=0.025 + 0.040 = 0.065\ \text{L}\)
\(n\ce{(NO3^-)}= 3 \times n(Fe^{3+}) = 3 \times 0.00375 = 0.01125\ \text{mol}\)
\(c\ce{(NO3^-)} = \dfrac{n}{V} = \dfrac{0.01125}{0.065} = 0.173\ \text{mol L}^{-1}\)
The compound potassium nitrate has the formula \(\ce{KNO3}\).
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a. \(1.58\ \text{mol L}^{-1}\)
b. \(0.156\ \text{L}\)
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.
\(n\ce{(KNO3)} = \dfrac{25}{101.11} = 0.247\ \text{mol}\)
\(c\ce{(KNO3)} = \dfrac{n}{V} = \dfrac{0.247}{0.5} = 0.494\ \text{mol L}^{-1}\)
| \(c_1V_1\) | \(=c_2V_2\) | |
| \(1.58 \times V_1\) | \(=0.494 \times 0.5\) | |
| \(V_1\) | \(=\dfrac{0.247}{1.58}\) | |
| \(V_1\) | \(=0.156\ \text{L}\) |
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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\(0.189\ \text{mol L}^{-1}\)
\(\ce{MM(Na2CO3)}= 105.99\ \text{g mol}^{-1}\)
\(\ce{n(Na2CO3)}= \dfrac{5}{105.99} = 0.0472\ \text{mol}\)
Concentration of \(\ce{Na2CO3}\):
\(c= \dfrac{\text{n}}{\text{V}} = \dfrac{0.0472}{0.250}=0.189\ \text{mol L}^{-1}\)
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?
\(A\)
→ \(\ce{MM(C6H12O6)} = (6 \times 12.01) + (12 \times 1.008) + (6 \times 16.00) = 180.156\ \text{g mol}^{-1}\)
→ \(\ce{n(C6H12O6)} = \dfrac{12.0 \times 10^{-3}}{180.156}=6.66 \times 10^{-5}\ \text{mol}\)
→ \(c=\dfrac{\text{n}}{\text{V}}=\dfrac{6.66 \times 10^{-5}}{50 \times 10^{-3}}=1.3 \times 10^{-3}\ \text{mol L}^{-1}\)
\(\Rightarrow A\)
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)
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\(\text{0.342 M}\)
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 \(0.342\ \text{M}\).
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) --- 5 WORK AREA LINES (style=lined) --- \(\ce{[H2O2] = 27.4\ \text{g L}^{-1}}\) \(\ce{n(O2) = \dfrac{10.0}{24.8} = 0.403\ \text{mol (at SLC)}}\) \(\ce{n(H2O2) = 2 \times n(O2) = 2 \times 0.403 = 0.806\ \text{mol}}\) \(\ce{MM(H2O2) = 2 \times 1.008 + 2 \times 16 = 34.0\ \text{g}}\) \(\ce{m(H2O2) = 0.806 \times 34.0 = 27.4\ \text{g}}\) \(\therefore \ce{[H2O2] = 27.4\ \text{g L}^{-1}}\)
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♦♦ Mean mark 38%.
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?
\(A\)
\(\ce{n(HCl)} = 0.100 \times 24.4 \times 10^{-3} = 2.44 \times 10^{-3}\ \text{mol}\)
\(\ce{n(NaX)}=2.44 \times 10^{-3}\ \text{mol in 20 mL sample.}\)
\(\ce{c(NaX)}=\dfrac{2.44 \times 10^{-3}}{20 \times 10^{-3}} =0.122\ \text{mol L}^{-1}\)
\(\ce{n(NaX)}=0.122 \times 0.1=0.0122\ \text{mol (in 100 mL sample)}\)
\(\ce{MM(NaX)}=\dfrac{1}{0.0122}=82\ \text{g mol}^{-1}\)
\(\Rightarrow A\)
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)
\(\ce{0.18 mol L^{-1} \text{(to 2 d.p.)}}\)
\(\ce{n(C6H8O7) = c \times V = 0.100 \times 0.0250 = 0.00250 moles}\)
\(\ce{NaOH : C6H8O7}\ \ \text{reaction ratio is}\ 3:1\)
\(\ce{n(NaOH in 41.50 mL) = 3 \times 0.00250 = 0.00750 moles}\)
\[\ce{[NaOH] = \frac{n}{V} = \frac{0.00750}{0.04150} = 0.18072… = 0.18 mol L^{-1} \text{(to 2 d.p.)}}\]