Here are the solutions to the questions. 4. Calculate the mass D of Potassium hydroxide. Step 1: Calculate the moles of nitric acid ($\text{HNO}_3$) used. $$ \text{Moles of HNO}_3 = \text{Concentration} \times \text{Volume} $$ $$ \text{Moles of HNO}_3 = 2.0 \text{ mol/L} \times \frac{50}{1000} \text{ L} = 0.1 \text{ mol} $$ Step 2: Determine the moles of potassium hydroxide ($\text{KOH}$) in 50 $\text{cm}^3$ of solution. The balanced equation is $\text{KOH}_{(aq)} + \text{HNO}_{3(aq)} \rightarrow \text{KNO}_{3(aq)} + \text{H}_2\text{O}_{(l)}$. From the equation, the mole ratio of $\text{KOH} : \text{HNO}_3$ is $1:1$. Therefore, moles of $\text{KOH}$ in 50 $\text{cm}^3$ solution = moles of $\text{HNO}_3$ = 0.1 mol. Step 3: Calculate the moles of $\text{KOH}$ in 100 $\text{cm}^3$ of solution. Since 50 $\text{cm}^3$ of solution contains 0.1 mol of $\text{KOH}$, then 100 $\text{cm}^3$ of solution contains: $$ \text{Moles of KOH in 100 cm}^3 = 0.1 \text{ mol} \times \frac{100 \text{ cm}^3}{50 \text{ cm}^3} = 0.2 \text{ mol} $$ Step 4: Calculate the mass D of $\text{KOH}$. Relative formula mass (RFM) of $\text{KOH} = 56 \text{ g/mol}$. $$ \text{Mass D} = \text{Moles} \times \text{RFM} $$ $$ \text{Mass D} = 0.2 \text{ mol} \times 56 \text{ g/mol} = 11.2 \text{ g} $$ The mass D of Potassium hydroxide is $\boxed{11.2 \text{ g}}$. 5. Calculate the mass of sodium sulphite that was used. Step 1: Write the balanced chemical equation for the reaction. $$ \text{Na}_2\text{SO}_3(s) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{SO}_2(g) $$ Step 2: Calculate the moles of sulphur dioxide ($\text{SO}_2$) gas produced. Molar gas volume = 24000 $\text{cm}^3$/mol. $$ \text{Moles of SO}_2 = \frac{\text{Volume of SO}_2}{\text{Molar gas volume}} $$ $$ \text{Moles of SO}_2 = \frac{960 \text{ cm}^3}{24000 \text{ cm}^3/\text{mol}} = 0.04 \text{ mol} $$ Step 3: Determine the moles of sodium sulphite ($\text{Na}_2\text{SO}_3$) used. From the balanced equation, the mole ratio of $\text{Na}_2\text{SO}_3 : \text{SO}_2$ is $1:1$. Therefore, moles of $\text{Na}_2\text{SO}_3$ used = 0.04 mol. Step 4: Calculate the mass of sodium sulphite used. Molar mass of sodium sulphite = 126 g/mol. $$ \text{Mass of Na}_2\text{SO}_3 = \text{Moles} \times \text{Molar mass} $$ $$ \text{Mass of Na}_2\text{SO}_3 = 0.04 \text{ mol} \times 126 \text{ g/mol} = 5.04 \text{ g} $$ The mass of sodium sulphite used is $\boxed{5.04 \text{ g}}$. 6. Calculate the volume of chlorine which will react with iron to form 0.5g of Iron (III) chloride. Step 1: Calculate the molar mass of Iron (III) chloride ($\text{FeCl}_3$). Atomic mass of Fe = 56 Atomic mass of Cl = 35.5 $$ \text{Molar mass of FeCl}_3 = 56 + (3 \times 35.5) = 56 + 106.5 = 162.5 \text{ g/mol} $$ Step 2: Calculate the moles of $\text{FeCl}_3$ formed. $$ \text{Moles of FeCl}_3 = \frac{\text{Mass of FeCl}_3}{\text{Molar mass of FeCl}_3} $$ $$ \text{Moles of FeCl}_3 = \frac{0.5 \text{ g}}{162.5 \text{ g/mol}} \approx 0.0030769 \text{ mol} $$ Step 3: Determine the moles of chlorine ($\text{Cl}_2$) required. The balanced equation is $2\text{Fe}_{(s)} + 3\text{Cl}_{2(g)} \rightarrow 2\text{FeCl}_{3(s)}$. From the equation, the mole ratio of $\text{Cl}_2 : \text{FeCl}_3$ is $3:2$. $$ \text{Moles of Cl}_2 = \frac{3}{2} \times \text{Moles of FeCl}_3 $$ $$ \text{Moles of Cl}_2 = \frac{3}{2} \times 0.0030769 \text{ mol} = 0.00461535 \text{ mol} $$ Step 4: Calculate the volume of chlorine gas. Molar gas volume at 298K = 24 $\text{dm}^3$/mol. $$ \text{Volume of Cl}_2 = \text{Moles of Cl}_2 \times \text{Molar gas volume} $$ $$ \text{Volume of Cl}_2 = 0.00461535 \text{ mol} \times 24 \text{ dm}^3/\text{mol} = 0.1107684 \text{ dm}^3 $$ Converting to $\text{cm}^3$: $$ 0.1107684 \text{ dm}^3 \times 1000 \text{ cm}^3/\text{dm}^3 = 110.7684 \text{ cm}^3 $$ The volume of chlorine required is approximately $\boxed{110.8 \text{ cm}^3}$. 7. Calculate the concentration of the solution in moles per litre. Step 1: Calculate the mass of ethanoic acid ($\text{CH}_3\text{COOH}$) used. Density of ethanoic acid = 1.05 $\text{g/cm}^3$. Volume of ethanoic acid = 15.0 $\text{cm}^3$. $$ \text{Mass of CH}_3\text{COOH} = \text{Density} \times \text{Volume} $$ $$ \text{Mass of CH}_3\text{COOH} = 1.05 \text{ g/cm}^3 \times 15.0 \text{ cm}^3 = 15.75 \text{ g} $$ Step 2: Calculate the molar mass of ethanoic acid ($\text{CH}_3\text{COOH}$). Atomic masses: C=12, H=1, O=16. $$ \text{Molar mass of CH}_3\text{COOH} = (2 \times 12) + (4 \times 1) + (2 \times 16) $$ $$ \text{Molar mass of CH}_3\text{COOH} = 24 + 4 + 32 = 60 \text{ g/mol} $$ Step 3: Calculate the moles of ethanoic acid. $$ \text{Moles of CH}_3\text{COOH} = \frac{\text{Mass of CH}_3\text{COOH}}{\text{Molar mass of CH}_3\text{COOH}} $$ $$ \text{Moles of CH}_3\text{COOH} = \frac{15.75 \text{ g}}{60 \text{ g/mol}} = 0.2625 \text{ mol} $$ Step 4: Calculate the concentration of the solution in moles per litre. Total volume of solution = 500 $\text{cm}^3 = 0.5 \text{ L}$. $$ \text{Concentration} = \frac{\text{Moles of CH}_3\text{COOH}}{\text{Volume of solution (L)}} $$ $$ \text{Concentration} = \frac{0.2625 \text{ mol}}{0.5 \text{ L}} = 0.525 \text{ mol/L} $$ The concentration of the solution is $\boxed{0.525 \text{ mol/L}}$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.