If 0.24 g of a volatile liquid upon vaporization gives 45 ml of vapours at NTP.What will be the vapour density of the substance ? (Density of `H_(2) =0.089 gL^(-1))`

Options

A. `95.39`
B. `5.993`
C. `95.93`
D. `56.0`

Explanation

To find the vapor density of the volatile liquid, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Given Data**:
- Mass of the volatile liquid (m) = 0.24 g
- Volume of vapor at NTP (V) = 45 ml
- At Normal Temperature and Pressure (NTP), 1 mole of gas occupies 22,400 ml.

2. **Calculate the Mass of Vapor for 22,400 ml**:
- We know that 0.24 g of the liquid produces 45 ml of vapor. We need to find out how much mass would correspond to 22,400 ml.
- Using the proportion:
\[ \text{Mass for 22,400 ml} = \frac{0.24 \, \text{g}}{45 \, \text{ml}} \times 22,400 \, \text{ml} \]
- Performing the calculation:
\[ \text{Mass for 22,400 ml} = \frac{0.24 \times 22,400}{45} = \frac{5376}{45} \approx 119.47 \, \text{g} \]

3. **Calculate the Molar Mass**:
- The mass calculated above (119.47 g) corresponds to 1 mole of the vapor since it is the mass for 22,400 ml at NTP.
- Therefore, the molar mass (M) of the substance is approximately 119.47 g/mol.

4. **Calculate the Vapor Density**:
- The vapor density (D) is defined as half of the molar mass:
\[ D = \frac{M}{2} = \frac{119.47}{2} \approx 59.735 \, \text{g/mol} \]
- Rounding off, we can say the vapor density is approximately 60 g/mol.

5. **Final Answer**:
- The vapor density of the substance is approximately 60 g/mol.

### Conclusion:
The closest answer from the options provided would be option 4, which is 56 g/mol.