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:
The number of n. bonds present in the compound A is/are

Options

A. 2
B. 1
C. 3
D. 4

Answer

To solve the problem step by step, we will use Graham's law of effusion and the information provided about the hydrocarbon (A) and its reaction with hydrogen gas.

### Step 1: Use Graham's Law of Effusion
Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

Given:
- Methane (CH₄) diffuses 1 g in 20 seconds.
- Hydrocarbon (A) diffuses √20 g in 40 seconds.

Let the molar mass of methane (CH₄) be M₁ and that of hydrocarbon A be M₂.

Using Graham's law:
\[ \frac{R_{CH_4}}{R_A} = \sqrt{\frac{M_A}{M_{CH_4}}} \]
Where \(R\) is the rate of diffusion.

### Step 2: Calculate the Rates of Diffusion
The rate of diffusion can be calculated as the mass diffused per unit time.

For methane:
\[ R_{CH_4} = \frac{1 \text{ g}}{20 \text{ s}} = 0.05 \text{ g/s} \]

For hydrocarbon A:
\[ R_A = \frac{\sqrt{20} \text{ g}}{40 \text{ s}} = \frac{4.47 \text{ g}}{40 \text{ s}} = 0.11175 \text{ g/s} \]

### Step 3: Set Up the Equation Using Graham's Law
Substituting the rates into Graham's law:
\[ \frac{0.05}{0.11175} = \sqrt{\frac{M_A}{16}} \quad \text{(since Molar mass of CH₄ = 16 g/mol)} \]

### Step 4: Solve for Molar Mass of Hydrocarbon A
Calculating the left side:
\[ \frac{0.05}{0.11175} \approx 0.447 \]
Now squaring both sides:
\[ 0.447^2 = \frac{M_A}{16} \]
\[ 0.199 = \frac{M_A}{16} \]
\[ M_A = 0.199 \times 16 = 3.184 \text{ g/mol} \]

### Step 5: Determine the Number of Bonds in 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 use the ideal gas law to find the moles of H₂.

Using the ideal gas law:
\[ PV = nRT \]
Where:
- P = 760 mmHg = 1 atm
- V = 8.40 mL = 0.0084 L
- R = 0.0821 L·atm/(K·mol)
- T = 273 K

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

### Step 6: Relate Moles of A to Moles of H₂
From the reaction of A with H₂, we can assume that 1 mole of A reacts with 1 mole of H₂ to form a saturated product.

Given that:
\[ \text{Mass of A} = 10 \text{ mg} = 0.01 \text{ g} \]
Calculating moles of A:
\[ \text{Moles of A} = \frac{0.01 \text{ g}}{3.184 \text{ g/mol}} \approx 0.00314 \text{ moles} \]

### Step 7: Determine the Number of π Bonds
If 1 mole of A gives 1 mole of saturated product, and we know the number of moles of H₂ reacted, we can determine the number of π bonds.

Since each π bond requires one H₂ molecule:
\[ \text{Number of π bonds} = \text{Moles of H₂} = 0.00037 \text{ moles} \]

### Conclusion
The number of π bonds in compound A is 3.