Question

Concentrated nitric acid is labelled as 75%by mass. The volume in mL of the solution which contains 30 g of nitric acid is: Given: Density of nitric acid solution is 1.25 g/mL.

• A. 55
• B. 45
• C. 32
• D. 40

Hint:

When solving mass and volume problems, use the relationship \( {Density} = \frac{{Mass}}{{Volume}} \) to find the missing values.

Answer 1

The Correct Option is C

To find the volume of the nitric acid solution that contains 30 g of nitric acid in a solution that is 75% nitric acid by mass, we can follow these steps: 

1. Calculate the total mass of the solution required to contain 30 g of nitric acid. Since the solution is 75% nitric acid by mass, the formula to find the total mass (m_total) is given by:
   
   \( m_{\text{HNO}_3} = \frac{75}{100} \times m_{\text{total}} \)
   
   Given that \( m_{\text{HNO}_3} = 30 \) g, we have:
   
   \( 30 = \frac{75}{100} \times m_{\text{total}} \)
   \( m_{\text{total}} = \frac{30 \times 100}{75} \)
   \( m_{\text{total}} = 40 \) g
2. Calculate the volume of the solution using its density. The density (\( \rho \)) of the nitric acid solution is given as 1.25 g/mL. We can use the formula:
   
   \( \text{Volume (mL)} = \frac{\text{Mass (g)}}{\text{Density (g/mL)}} \)
   
   Substituting the values:
   
   \( \text{Volume} = \frac{40}{1.25} \)
   \( \text{Volume} = 32 \) mL

Thus, the volume of the solution required is 32 mL.

Answer 2

Step 1: Given data.
Mass percentage of nitric acid = 75% by mass
Density of nitric acid solution = 1.25 g/mL
Mass of pure HNO₃ required = 30 g

Step 2: Relation between mass of solution and mass percentage.
If the solution is 75% by mass, it means:
\[ 75 \, \text{g of HNO₃ is present in 100 g of solution.} \]
Therefore, to get 30 g of HNO₃, we need:
\[ \text{Mass of solution} = \frac{30 \times 100}{75} = 40 \, \text{g}. \]

Step 3: Find the volume of the solution.
Using the density formula:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \Rightarrow \text{Volume} = \frac{\text{Mass}}{\text{Density}}. \]
\[ \text{Volume} = \frac{40}{1.25} = 32 \, \text{mL}. \]

Final Answer:
\[ \boxed{32 \, \text{mL}} \]