The molecular weights of two ideal gases A and B are respectively 100 and 200. One gram of A occupies V litres of volume at STP. What is the volume (in litres) occupied by one gram of B at STP ?

Options

A. V/2
B. V
C. 4V
D. 2V

Answer

To solve the problem, we need to determine the volume occupied by 1 gram of gas B at STP, given that 1 gram of gas A occupies V liters at STP. We will use the concept of moles and Avogadro's law.

### Step-by-Step Solution:

1. **Calculate the number of moles of gas A:**
\[ \text{Number of moles of A} = \frac{\text{Given mass of A}}{\text{Molar mass of A}} = \frac{1 \text{ g}}{100 \text{ g/mol}} = 0.01 \text{ moles} \]

2. **Calculate the number of moles of gas B:**
\[ \text{Number of moles of B} = \frac{\text{Given mass of B}}{\text{Molar mass of B}} = \frac{1 \text{ g}}{200 \text{ g/mol}} = 0.005 \text{ moles} \]

3. **Apply Avogadro's law:**
According to Avogadro's law, at the same temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas. Therefore, we can set up a ratio:
\[ \frac{V_A}{V_B} = \frac{n_A}{n_B} \]
where \( V_A \) is the volume occupied by gas A, \( V_B \) is the volume occupied by gas B, \( n_A \) is the number of moles of gas A, and \( n_B \) is the number of moles of gas B.

4. **Substituting the values:**
\[ \frac{V}{V_B} = \frac{0.01}{0.005} \]
Simplifying this gives:
\[ \frac{V}{V_B} = 2 \implies V_B = \frac{V}{2} \]

5. **Conclusion:**
The volume occupied by 1 gram of gas B at STP is:
\[ V_B = \frac{V}{2} \]

### Final Answer:
The volume occupied by 1 gram of gas B at STP is \( \frac{V}{2} \) liters.