Question

0.24 g of a volatile gas, upon vaporisation, gives 45 mL vapour at NTP. What will be the vapour density of the substance ? (Density of $ H_2 = 0.089$)

• A. 95.93
• B. 59.93
• C. 95.39
• D. 5.993

Answer 1

The Correct Option is B

To find the vapour density of the volatile gas, we will use the known relation between the mass, volume, and molar mass of a gas at Normal Temperature and Pressure (NTP). Here, the steps are as follows:

1. At NTP, 1 mole of any ideal gas occupies 22.4 L.
2. Given that 0.24 g of the gas occupies 45 mL (or 0.045 L) at NTP, we can find the molar mass by using the proportionality of these volumes.
3. Using the proportion: \frac{\text{Molar mass of gas}}{\text{Mass of gas}} = \frac{\text{Volume of 1 mole of gas at NTP}}{\text{Volume occupied by gas}} = \frac{22.4 \text{ L}}{0.045 \text{ L}}.

Let's calculate the molar mass:

\text{Molar mass} = 0.24 \, \text{g} \times \frac{22.4}{0.045} = 0.24 \times 497.78 = 119.47 \, \text{g/mol}

Once we have the molar mass, we can find the vapour density. Vapour density is defined as the mass of a certain volume of the gas divided by the mass of the same volume of hydrogen gas (at the same temperature and pressure). Mathematically:

\text{Vapour Density} = \frac{\text{Molar mass of the gas}}{2} = \frac{119.47}{2} \approx 59.735 \, \text{(rounding to two decimal points, 59.73)}

Among the given options, the closest match to our calculated vapour density (rounded to two decimal places) is 59.93, which is the correct answer.