If V ml. of the vapours of substance at STP weight W g. Then molecular weight of substance is :-

Options

A. `(W//V)xx22400`
B. `(V)/(W)xx22.4`
C. `(W-V)xx22400`
D. `(Wxx1)/(Vxx22400)`

Explanation

To find the molecular weight of a substance given its vapor volume at STP and weight, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Volume of vapor = V ml
- Weight of vapor = W g
- Conditions = STP (Standard Temperature and Pressure)

2. **Convert Volume from ml to Liters**:
- Since 1 liter = 1000 ml, we convert V ml to liters:
\[ V \text{ (in liters)} = \frac{V}{1000} \]

3. **Use the Ideal Gas Law**:
- The Ideal Gas Law is given by:
\[ PV = nRT \]
- Where:
- \( P \) = pressure (1 bar at STP)
- \( V \) = volume in liters
- \( n \) = number of moles
- \( R \) = universal gas constant (0.0831 L·bar/(K·mol))
- \( T \) = temperature (273 K at STP)

4. **Relate Moles to Weight and Molecular Weight**:
- The number of moles \( n \) can be expressed as:
\[ n = \frac{W}{M} \]
- Where \( M \) is the molecular weight we want to find.

5. **Substitute into the Ideal Gas Law**:
- Substitute \( n \) into the Ideal Gas Law:
\[ PV = \frac{W}{M} RT \]

6. **Rearranging the Equation**:
- Rearranging the equation to solve for molecular weight \( M \):
\[ M = \frac{WRT}{PV} \]

7. **Substituting Known Values**:
- Substitute \( R = 0.0831 \, \text{L·bar/(K·mol)} \), \( T = 273 \, \text{K} \), \( P = 1 \, \text{bar} \), and \( V = \frac{V}{1000} \):
\[ M = \frac{W \times 0.0831 \times 273}{1 \times \frac{V}{1000}} \]

8. **Simplifying the Expression**:
- Simplifying gives:
\[ M = \frac{W \times 0.0831 \times 273 \times 1000}{V} \]
- Calculate \( 0.0831 \times 273 \times 1000 \):
\[ 0.0831 \times 273 \times 1000 = 22700.3 \approx 22400 \]
- Thus, the molecular weight becomes:
\[ M \approx \frac{22400 \times W}{V} \]

### Final Answer:
The molecular weight of the substance is:
\[ M = \frac{22400 \times W}{V} \]