Question

Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL^–1 and the mass per cent of nitric acid in it being 69%.

Answer 1

Given data:

• Density of solution = \(1.41 \,\text{g mL}^{-1}\)
• Mass percent of HNO\(_3\) = \(69\%\)

Step 1: Mass of HNO\(_3\) in 100 g solution

Assume \(100 \,\text{g}\) of solution.

Mass of HNO\(_3\) in 100 g solution = \(69 \,\text{g}\).

Step 2: Moles of HNO\(_3\)

Molar mass of HNO\(_3\) = \(1 + 14 + 3 \times 16 = 63 \,\text{g mol}^{-1}\).

Number of moles of HNO\(_3\) in \(69 \,\text{g}\):

\[ n = \frac{69}{63} \approx 1.095 \,\text{mol} \]

Step 3: Volume of 100 g solution

Density \( \rho = \dfrac{\text{mass}}{\text{volume}} \Rightarrow \text{volume} = \dfrac{\text{mass}}{\rho} \).

\[ V = \frac{100 \,\text{g}}{1.41 \,\text{g mL}^{-1}} \approx 70.92 \,\text{mL} \]

\[ 70.92 \,\text{mL} = 70.92 \times 10^{-3} \,\text{L} = 0.07092 \,\text{L} \]

Step 4: Concentration (Molarity)

Molarity \(M\) is given by \( M = \dfrac{\text{moles of solute}}{\text{volume of solution in L}} \).

\[ M = \frac{1.095}{0.07092} \approx 15.4 \,\text{mol L}^{-1} \]

Final Answer

The concentration of nitric acid in the sample is approximately 15.4 mol L\(^{-1}\).