In a Victor Meyer's determination, 0.23 g of a volatile substance displaced air which measured 112 mL at S.T.P. Calculate the vapour density and molecular weight of the substance (1 litre of `H_2` at S.T.P. weighs 0.09 g).

Explanation

To solve the problem of determining the vapor density and molecular weight of a volatile substance using Victor Meyer's method, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data:**
- Mass of the volatile substance (m) = 0.23 g
- Volume of air displaced (V) = 112 mL
- At STP (Standard Temperature and Pressure), 1 L of H₂ weighs 0.09 g.

2. **Convert Volume from mL to L:**
- Since 1 L = 1000 mL, we convert 112 mL to liters:
\[ V = \frac{112 \, \text{mL}}{1000} = 0.112 \, \text{L} \]

3. **Use the Ideal Gas Equation:**
- The ideal gas equation can be rearranged to find the molecular weight (M):
\[ PV = \frac{m}{M}RT \]
- Where:
- P = Pressure (1 atm)
- V = Volume (0.112 L)
- m = Mass of the substance (0.23 g)
- R = Ideal gas constant (0.0821 L·atm/(K·mol))
- T = Temperature (273 K)

4. **Rearranging the Equation:**
- Rearranging the equation for M gives:
\[ M = \frac{mRT}{PV} \]

5. **Substituting Values:**
- Substitute the known values into the equation:
\[ M = \frac{0.23 \, \text{g} \times 0.0821 \, \text{L·atm/(K·mol)} \times 273 \, \text{K}}{1 \, \text{atm} \times 0.112 \, \text{L}} \]

6. **Calculating Molecular Weight:**
- Calculate the numerator:
\[ 0.23 \times 0.0821 \times 273 = 5.173 \, \text{g·L·atm/(K·mol)} \]
- Now divide by the volume:
\[ M = \frac{5.173}{0.112} \approx 46.02 \, \text{g/mol} \]

7. **Calculate Vapor Density (D):**
- Vapor density is calculated using the formula:
\[ D = \frac{M}{2} \]
- Substitute the molecular weight:
\[ D = \frac{46.02}{2} = 23.01 \, \text{g/L} \]

### Final Answers:
- Molecular Weight (M) = 46.02 g/mol
- Vapor Density (D) = 23.01 g/L