When 4 g of an ideal gas A is introduced into an evacuated flask kept at `25^(@)C`, the pressure is found to be one atmosphere. If 6 g of another ideal gas B is then added to the same flask, the pressure becomes 2 atm at same temperature. The ratio of molecular weights `(M_(A) : M_(B))` of the two gases would be

Options

A. `1:2`
B. `2:1`
C. `2:3`
D. `3:2`

Explanation

To find the ratio of molecular weights \( (M_A : M_B) \) of the two gases A and B, we can use the ideal gas law and the information given in the problem.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have two gases: A and B.
- 4 g of gas A is introduced into an evacuated flask at 25°C (298 K), resulting in a pressure of 1 atm.
- Then, 6 g of gas B is added, raising the pressure to 2 atm.

2. **Using the Ideal Gas Law:**
The ideal gas law is given by:
\[ PV = nRT \]
where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin

3. **Calculating Moles of Gas A:**
For gas A, when the pressure is 1 atm:
\[ P_A = 1 \text{ atm}, \quad T = 298 \text{ K} \]
Rearranging the ideal gas law to find the number of moles \( n_A \):
\[ n_A = \frac{P_A V}{RT} \]
Since we don’t know \( V \), we will keep it as a variable.

4. **Calculating Moles of Gas B:**
When gas B is added, the total pressure becomes 2 atm:
\[ P_{total} = P_A + P_B = 2 \text{ atm} \]
This means the pressure contributed by gas B is also 1 atm (since \( P_A = 1 \text{ atm} \)).
\[ P_B = 1 \text{ atm} \]
The number of moles of gas B can be calculated similarly:
\[ n_B = \frac{P_B V}{RT} \]

5. **Relating Moles to Mass and Molar Mass:**
The number of moles can also be expressed in terms of mass and molar mass:
\[ n_A = \frac{W_A}{M_A} \quad \text{and} \quad n_B = \frac{W_B}{M_B} \]
where \( W_A = 4 \text{ g} \) and \( W_B = 6 \text{ g} \).

6. **Setting Up the Equation:**
Since \( n_A = n_B \) (because both pressures are equal and the volume and temperature are constant), we can set the equations equal:
\[ \frac{W_A}{M_A} = \frac{W_B}{M_B} \]

7. **Substituting the Known Values:**
Substituting \( W_A \) and \( W_B \):
\[ \frac{4 \text{ g}}{M_A} = \frac{6 \text{ g}}{M_B} \]

8. **Finding the Ratio of Molar Masses:**
Rearranging gives:
\[ \frac{M_A}{M_B} = \frac{4}{6} = \frac{2}{3} \]

9. **Final Answer:**
The ratio of the molecular weights of gases A and B is:
\[ M_A : M_B = 2 : 3 \]