When `2g` of gas `A` is introduced into an evacuated flask kept at `25^(@)C` the presusre is found to be `1atm`. If `3g` of another gas `B` is then added to the same flask the total pressure becomes `1.5 atm`. The ratio of molecular weights M A : M B .

Explanation

To solve the problem, we will use the ideal gas law and the information provided in the question. Here’s a step-by-step solution:

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
where:
- \( P \) = pressure,
- \( V \) = volume,
- \( n \) = number of moles,
- \( R \) = universal gas constant,
- \( T \) = temperature in Kelvin.

### Step 2: Calculate Moles of Gas A
When 2 g of gas A is introduced into the flask, the pressure is 1 atm. We can express the number of moles \( n_A \) of gas A as:
\[ n_A = \frac{w_A}{M_A} \]
where \( w_A = 2 \, \text{g} \) and \( M_A \) is the molar mass of gas A.

Using the ideal gas law for gas A:
\[ P_A V = n_A RT \]
Substituting for \( n_A \):
\[ P_A V = \frac{w_A}{M_A} RT \]
Rearranging gives:
\[ M_A = \frac{w_A RT}{P_A V} \]

### Step 3: Substitute Known Values for Gas A
We know:
- \( w_A = 2 \, \text{g} \),
- \( P_A = 1 \, \text{atm} \),
- \( R \) (universal gas constant) is approximately \( 0.0821 \, \text{L atm/(K mol)} \),
- \( T = 25^\circ C = 298 \, \text{K} \).

Since the volume \( V \) is constant, we can express it in terms of the pressure and other variables. Thus:
\[ M_A = \frac{2 \times 0.0821 \times 298}{1 \times V} \]

### Step 4: Calculate Moles of Gas B
When 3 g of gas B is added, the total pressure becomes 1.5 atm. The pressure of gas B can be calculated as:
\[ P_{mix} = P_A + P_B \]
Given \( P_{mix} = 1.5 \, \text{atm} \) and \( P_A = 1 \, \text{atm} \):
\[ P_B = 1.5 - 1 = 0.5 \, \text{atm} \]

Now, for gas B:
\[ n_B = \frac{w_B}{M_B} \]
where \( w_B = 3 \, \text{g} \).

Using the ideal gas law for gas B:
\[ P_B V = n_B RT \]
Substituting for \( n_B \):
\[ P_B V = \frac{w_B}{M_B} RT \]
Rearranging gives:
\[ M_B = \frac{w_B RT}{P_B V} \]

### Step 5: Substitute Known Values for Gas B
We know:
- \( w_B = 3 \, \text{g} \),
- \( P_B = 0.5 \, \text{atm} \).

Thus:
\[ M_B = \frac{3 \times 0.0821 \times 298}{0.5 \times V} \]

### Step 6: Find the Ratio of Molecular Weights
Now we can find the ratio \( \frac{M_A}{M_B} \):
\[ \frac{M_A}{M_B} = \frac{\frac{2 \times 0.0821 \times 298}{1 \times V}}{\frac{3 \times 0.0821 \times 298}{0.5 \times V}} \]
The \( R \) and \( V \) cancel out:
\[ \frac{M_A}{M_B} = \frac{2}{3 \times 0.5} = \frac{2}{1.5} = \frac{4}{3} \]

### Final Answer
The ratio of molecular weights \( M_A : M_B = 4 : 3 \).

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