1 g of methane diffused in 20 sec. under certain conditions. Under the same conditions `sqrt(20)g` of a hydrocarbon (A) diffused in 40 sec. A 10 mg of sample of (A) took up 8.40 ml of `H_(2)` gas measured at 0°C and 760 mm pressure:
Identify the incorrect statement.

Options

A. One of the isomers of (A) exhibits geometrical isomerism
B. One of the isomers of (A) gives white precipitate with Tollen's reagent
C. One of the isomers of (A) is asymmetric
D. One of the isomers on ozonolysis gives only acetone

Answer

To solve the problem, we need to analyze the information given about the diffusion of methane and hydrocarbon A, and then evaluate the statements provided. Here’s a step-by-step breakdown:

### Step 1: Understand the diffusion of gases
According to Graham's law of effusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. This can be expressed as:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
where \( R \) is the rate of diffusion and \( M \) is the molar mass of the gases.

### Step 2: Calculate the molar mass of methane (CH₄)
The molar mass of methane (CH₄) is:
\[ M_{CH_4} = 12 + (4 \times 1) = 16 \text{ g/mol} \]

### Step 3: Set up the diffusion rates
Given:
- 1 g of methane diffuses in 20 seconds.
- The mass of hydrocarbon A that diffuses is \( \sqrt{20} \) g in 40 seconds.

Let’s denote the molar mass of hydrocarbon A as \( M_A \).

### Step 4: Calculate the rate of diffusion for both gases
Using the information provided:
- Rate of diffusion of methane:
\[ R_{CH_4} = \frac{1 \text{ g}}{20 \text{ s}} = 0.05 \text{ g/s} \]
- Rate of diffusion of hydrocarbon A:
\[ R_A = \frac{\sqrt{20} \text{ g}}{40 \text{ s}} = \frac{\sqrt{20}}{40} \text{ g/s} \]

### Step 5: Apply Graham's law
Using Graham's law:
\[ \frac{R_{CH_4}}{R_A} = \sqrt{\frac{M_A}{M_{CH_4}}} \]
Substituting the known values:
\[ \frac{0.05}{\frac{\sqrt{20}}{40}} = \sqrt{\frac{M_A}{16}} \]
This simplifies to:
\[ \frac{0.05 \times 40}{\sqrt{20}} = \sqrt{\frac{M_A}{16}} \]
Calculating the left side:
\[ \frac{2}{\sqrt{20}} = \sqrt{\frac{M_A}{16}} \]

### Step 6: Solve for \( M_A \)
Squaring both sides:
\[ \frac{4}{20} = \frac{M_A}{16} \]
\[ M_A = \frac{4 \times 16}{20} = 3.2 \text{ g/mol} \]

### Step 7: Analyze the properties of hydrocarbon A
Given that 10 mg of sample A took up 8.40 mL of H₂ gas at 0°C and 760 mm pressure, we can find the number of moles of A:
Using the ideal gas law:
\[ PV = nRT \]
Where:
- \( P = 760 \text{ mmHg} = 1 \text{ atm} \)
- \( V = 8.40 \text{ mL} = 0.0084 \text{ L} \)
- \( R = 0.0821 \text{ L atm/(K mol)} \)
- \( T = 273 \text{ K} \)

Calculating moles of H₂:
\[ n = \frac{PV}{RT} = \frac{(1)(0.0084)}{(0.0821)(273)} \approx 0.00037 \text{ moles} \]

### Step 8: Evaluate the statements
Now we evaluate the statements given:
1. One of the isomers of A exhibits geometrical isomerism.
2. One of the isomers of A gives a white precipitate with Tollens' reagent.
3. One of the isomers of A is asymmetric.
4. One of the isomers of A ozonolyses to give only acetone.

### Step 9: Identify the incorrect statement
- Statement 1 is true; hydrocarbons with double bonds can exhibit geometrical isomerism.
- Statement 2 is also true; terminal alkynes can give a white precipitate with Tollens' reagent.
- Statement 3 is true; if there is a chiral center, it can be asymmetric.
- Statement 4 is incorrect; ozonolysis of hydrocarbons does not necessarily yield only acetone.

### Conclusion
The incorrect statement is **Option D**: "One of the isomers of A ozonolyses to give only acetone."

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