40 miligram diatomic volatile substance `(X_2)` is converted to vapour that displaced `4.92mL` of air at 1atm and 300k. Atomic mass of element X is nearly :

Options

A. 400
B. 240
C. 200
D. 100

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the mass of the diatomic substance to grams
The mass of the diatomic volatile substance \(X_2\) is given as 40 milligrams. We need to convert this to grams:
\[ \text{Mass} = 40 \text{ mg} = 40 \times 10^{-3} \text{ g} = 0.040 \text{ g} \]

### Step 2: Use the ideal gas law to find the number of moles
We know the volume of air displaced is 4.92 mL. We need to convert this to liters:
\[ \text{Volume} = 4.92 \text{ mL} = 4.92 \times 10^{-3} \text{ L} \]

Using the ideal gas law \(PV = nRT\), we can rearrange it to find the number of moles \(n\):
\[ n = \frac{PV}{RT} \]
Where:
- \(P = 1 \text{ atm}\)
- \(V = 4.92 \times 10^{-3} \text{ L}\)
- \(R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}\)
- \(T = 300 \text{ K}\)

Substituting the values:
\[ n = \frac{(1 \text{ atm})(4.92 \times 10^{-3} \text{ L})}{(0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1})(300 \text{ K})} \]

### Step 3: Calculate the number of moles
Calculating the denominator:
\[ n = \frac{4.92 \times 10^{-3}}{0.0821 \times 300} = \frac{4.92 \times 10^{-3}}{24.63} \approx 0.000199 \text{ mol} \]

### Step 4: Relate moles to molar mass
We know that the mass of the substance is 0.040 g and we have calculated the number of moles. The molar mass \(M\) can be calculated using the formula:
\[ M = \frac{\text{mass}}{n} \]
Substituting the values:
\[ M = \frac{0.040 \text{ g}}{0.000199 \text{ mol}} \approx 201.01 \text{ g/mol} \]

### Step 5: Calculate the atomic mass of element X
Since the substance is diatomic (\(X_2\)), the molar mass \(M\) is related to the atomic mass \(m\) of element \(X\) by:
\[ M = 2m \]
Thus, we can find the atomic mass \(m\):
\[ m = \frac{M}{2} = \frac{201.01}{2} \approx 100.5 \text{ g/mol} \]

### Final Answer
The atomic mass of element \(X\) is approximately **100.5 g/mol**.

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