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Q1.4:
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL–1?

Solution :

Step 1: Identifying Given Parameters

To determine the molarity of the solution, we extract the following data from the problem statement:

  • Mass percentage of $HNO_3$ = $68\%$ (by mass)
  • Density of the solution ($d$) = $1.504\text{ g mL}^{-1}$
  • Molar mass of Nitric Acid ($HNO_3$):
    Calculated as: $H(1) + N(14) + O(3 \times 16) = 1 + 14 + 48 = 63\text{ g mol}^{-1}$ [By Atomic Mass Addition Rule]

Step 2: Defining Basis for Calculation

Assume the total mass of the solution is $100\text{ g}$.

  • Mass of solute ($HNO_3$) = $68\text{ g}$
  • Mass of solution = $100\text{ g}$

Step 3: Calculating Volume of the Solution

Using the definition of density, where $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$ [Per definition of mass density]:

$V = \frac{\text{Mass of solution}}{\text{Density}} = \frac{100\text{ g}}{1.504\text{ g mL}^{-1}}$

$V = 66.49\text{ mL} = 0.06649\text{ L}$

Step 4: Calculating Moles of Solute

Using the relation $\text{Moles} (n) = \frac{\text{Mass}}{\text{Molar Mass}}$:

$n(HNO_3) = \frac{68\text{ g}}{63\text{ g mol}^{-1}} \approx 1.0794\text{ mol}$

Step 5: Final Molarity Calculation

Molarity ($M$) is defined as the number of moles of solute per liter of solution:

$M = \frac{n(HNO_3)}{V(\text{in Liters})}$

$M = \frac{1.0794\text{ mol}}{0.06649\text{ L}}$

$M \approx 16.234\text{ mol L}^{-1}$

Visual Representation of Concentration Composition

68% HNO3 (68g) 32% H2O (32g) Figure: Composition of 100g of 68% Nitric Acid Solution

Final Solution: The molarity of the concentrated nitric acid solution is $16.23\text{ M}$.


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