Question 6 of 36intermediate🔧 ApplyNumerical3 marks

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%.

Correct Answer

15.44 mol L−1^{-1}

Exercise: EXERCISES | Q: 1.6 | (Page 25)
For More Understanding

Explanation

To find the concentration in moles per litre (molarity), we first determine the mass of the solute (nitric acid) in a given mass of the solution using the mass per cent. Then, we find the volume of the solution using its density. Finally, we calculate the number of moles of the solute and divide it by the volume of the solution in litres.

Solution Steps

  1. Step 1: Assume a mass of solution and find the mass of nitric acid. Let the mass of the solution be 100 g. Given mass per cent of nitric acid = 69%. Mass of nitric acid in 100 g solution = 69 g.

  2. Step 2: Calculate the number of moles of nitric acid. Molar mass of nitric acid (HNO3HNO_3) = 1+14+(3×16)=631 + 14 + (3 \times 16) = 63 g mol−1^{-1}. Number of moles of HNO3HNO_3 = Mass of HNO3Molar mass of HNO3=69 g63 g mol−1=1.095\frac{\text{Mass of } HNO_3}{\text{Molar mass of } HNO_3} = \frac{69 \text{ g}}{63 \text{ g mol}^{-1}} = 1.095 mol.

  3. Step 3: Calculate the volume of the solution. Density of the solution = 1.41 g mL−1^{-1}. Volume of 100 g solution = MassDensity=100 g1.41 g mL−1=70.92\frac{\text{Mass}}{\text{Density}} = \frac{100 \text{ g}}{1.41 \text{ g mL}^{-1}} = 70.92 mL. Convert volume to litres: 70.921000\frac{70.92}{1000} L = 0.07092 L.

  4. Step 4: Calculate the molarity (concentration in moles per litre). Molarity = No. of moles of soluteVolume of solution in litres\frac{\text{No. of moles of solute}}{\text{Volume of solution in litres}} Molarity = 1.095 mol0.07092 L=15.44\frac{1.095 \text{ mol}}{0.07092 \text{ L}} = 15.44 mol L−1^{-1}.

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