Concentrated nitric acid used as a laboratory reagent is usually 69% by mass of nitric acid. Calculate the volume of the solution which contained 23 g of `HNO_3`.Density of cone. `HNO_3` solution is `1.41 g cm^(-3)`?

Answer

To solve the problem, we need to calculate the volume of a concentrated nitric acid solution that contains 23 grams of HNO₃, given that the solution is 69% by mass of nitric acid and has a density of 1.41 g/cm³.

### Step-by-Step Solution:

**Step 1: Determine the mass of the solution.**
We know that the mass percentage of nitric acid in the solution is given by the formula:

\[ \text{Mass \%} = \left( \frac{\text{Mass of solute}}{\text{Mass of solution}} \right) \times 100 \]

Given:
- Mass of HNO₃ (solute) = 23 g
- Mass percentage of HNO₃ = 69%

We can rearrange the formula to find the mass of the solution (M):

\[ 69 = \left( \frac{23}{M} \right) \times 100 \]

Now, solving for M:

\[ M = \frac{23 \times 100}{69} \]

Calculating this gives:

\[ M \approx 33.33 \text{ g} \]

**Step 2: Calculate the volume of the solution using its density.**
We know the density (d) of the solution is given as 1.41 g/cm³. The formula for density is:

\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]

Rearranging this to find the volume (V):

\[ V = \frac{\text{Mass}}{\text{Density}} \]

Substituting the values we have:

\[ V = \frac{33.33 \text{ g}}{1.41 \text{ g/cm}^3} \]

Calculating this gives:

\[ V \approx 23.47 \text{ cm}^3 \]

Since 1 cm³ = 1 mL, we can also express this as:

\[ V \approx 23.47 \text{ mL} \]

### Final Answer:
The volume of the concentrated nitric acid solution that contains 23 g of HNO₃ is approximately **23.47 mL**.

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