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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 Variables and Definitions

To determine the molarity ($M$) of the solution, we define the parameters provided:

  • Mass percentage of $\text{HNO}_3 = 68\%$ (w/w). This implies that $100\text{ g}$ of the solution contains $68\text{ g}$ of $\text{HNO}_3$.
  • Density of the solution ($\rho$) $= 1.504\text{ g mL}^{-1}$.
  • Molar mass of Nitric Acid ($\text{HNO}_3$):
    • $H = 1.008\text{ g mol}^{-1}$
    • $N = 14.007\text{ g mol}^{-1}$
    • $O = 15.999\text{ g mol}^{-1} \times 3 = 47.997\text{ g mol}^{-1}$
    • Total Molar Mass ($M_m$) $\approx 63.0\text{ g mol}^{-1}$ [Based on Atomic Mass summation].

Step 2: Calculating the Volume of the Solution

Given the mass of the solution is $100\text{ g}$, we calculate the volume ($V$) using the density relationship [Per the definition of density: $\rho = m/V$]:

$V = \frac{\text{Mass}}{\text{Density}} = \frac{100\text{ g}}{1.504\text{ g mL}^{-1}} = 66.49\text{ mL}$

Converting to Liters ($L$) for the molarity formula: $V = 0.06649\text{ L}$.

Step 3: Calculating Moles of Solute ($\text{HNO}_3$)

Using the mass of $\text{HNO}_3$ derived from the mass percentage [Per the mole concept: $n = \text{mass} / \text{molar mass}$]:

$n_{\text{HNO}_3} = \frac{68\text{ g}}{63.0\text{ g mol}^{-1}} \approx 1.0794\text{ mol}$.

Step 4: Calculation of Molarity ($M$)

Molarity is defined as the number of moles of solute per liter of solution [Per the IUPAC definition of Molarity]:

$M = \frac{n_{\text{solute}}}{V_{\text{solution (L)}}} = \frac{1.0794\text{ mol}}{0.06649\text{ L}}$

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

Visual Representation of Concentration

Solution Container (100g = 66.49 mL) Solute: 68g HNO3 (1.079 mol) Solvent: 32g H2O Resulting Molarity: 16.23 M

Final Solution: The molarity of the 68% nitric acid solution is 16.23 M.


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