To a given container having a pore of definite size, gas A `(mol. wt. 81)` is filled till the final pressure become `10 atm`. It was seen that in `50 "minutes" 10 g` of A was effused out. Now the container was compelety evacuated and filled with gas `B(mol. wt. 100)` till the final pressure becomes `20 atm`. In `75` minuter how many gram of `B` will be effused out?

Options

A. `(100)/(6)g`
B. `(100)/(3)g`
C. `(200)/(3)g`
D. `(250)/(3)g`

Explanation

To solve the problem, we will use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The equation for Graham's law can be expressed as:

\[ \frac{W_A}{W_B} = \frac{P_B \cdot T_A}{P_A \cdot T_B} \cdot \sqrt{\frac{M_B}{M_A}} \]

Where:
- \(W_A\) = mass of gas A effused
- \(W_B\) = mass of gas B effused
- \(P_A\) = pressure of gas A
- \(P_B\) = pressure of gas B
- \(T_A\) = time for gas A
- \(T_B\) = time for gas B
- \(M_A\) = molar mass of gas A
- \(M_B\) = molar mass of gas B

### Step-by-Step Solution:

1. **Identify the given values for gas A:**
- Molar mass of gas A, \(M_A = 81 \, \text{g/mol}\)
- Pressure of gas A, \(P_A = 10 \, \text{atm}\)
- Time for gas A, \(T_A = 50 \, \text{minutes}\)
- Mass of gas A effused, \(W_A = 10 \, \text{g}\)

2. **Identify the given values for gas B:**
- Molar mass of gas B, \(M_B = 100 \, \text{g/mol}\)
- Pressure of gas B, \(P_B = 20 \, \text{atm}\)
- Time for gas B, \(T_B = 75 \, \text{minutes}\)

3. **Set up the equation using Graham's law:**
\[ \frac{10 \, \text{g}}{W_B} = \frac{20 \, \text{atm} \cdot 50 \, \text{minutes}}{10 \, \text{atm} \cdot 75 \, \text{minutes}} \cdot \sqrt{\frac{100 \, \text{g/mol}}{81 \, \text{g/mol}}} \]

4. **Calculate the right-hand side:**
- Calculate the pressure and time ratio:
\[ \frac{20 \cdot 50}{10 \cdot 75} = \frac{1000}{750} = \frac{4}{3} \]

- Calculate the square root of the molar mass ratio:
\[ \sqrt{\frac{100}{81}} = \frac{10}{9} \]

- Combine these results:
\[ \frac{4}{3} \cdot \frac{10}{9} = \frac{40}{27} \]

5. **Substitute back into the equation:**
\[ \frac{10}{W_B} = \frac{40}{27} \]

6. **Cross-multiply to solve for \(W_B\):**
\[ 10 \cdot 27 = 40 \cdot W_B \implies 270 = 40 \cdot W_B \implies W_B = \frac{270}{40} = 6.75 \, \text{g} \]

### Final Answer:
The mass of gas B that will be effused out is **6.75 grams**.