2g of a gas X are introduced into an evacuated flask kept at `25^(@)C`. The pressure is found to be 1 atm. If 3g of another gas Y are added to the same flask, the total pressure becomes 1.5 atm. Assuming that ideal behaviour, the molecular mass ratio of `M_(x)` and `M_(y)` is

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Initial Conditions
We have 2 g of gas X in a flask at a temperature of 25°C, and the initial pressure (P_X) is 1 atm.

### Step 2: Use the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

Since we are not given the volume, we can express everything in terms of the number of moles.

### Step 3: Calculate Moles of Gas X
Using the formula for moles:
\[ n = \frac{mass}{molar\ mass} \]
Let \( M_X \) be the molar mass of gas X. The number of moles of gas X is:
\[ n_X = \frac{2 \, \text{g}}{M_X} \]

### Step 4: Set Up the Equation for Gas X
From the ideal gas law for gas X:
\[ P_X V = n_X R T \]
Substituting \( n_X \):
\[ 1 \, \text{atm} \cdot V = \left(\frac{2}{M_X}\right) R (298) \]
Where 298 K is the temperature in Kelvin (25°C).

### Step 5: Introduce Gas Y
When 3 g of gas Y is added, the total pressure becomes 1.5 atm. Let \( M_Y \) be the molar mass of gas Y. The number of moles of gas Y is:
\[ n_Y = \frac{3 \, \text{g}}{M_Y} \]

### Step 6: Set Up the Equation for Total Pressure
The total pressure (P_total) in the flask is the sum of the partial pressures of gases X and Y:
\[ P_{total} = P_X + P_Y \]
Thus:
\[ 1.5 \, \text{atm} = 1 \, \text{atm} + P_Y \]
This gives:
\[ P_Y = 0.5 \, \text{atm} \]

### Step 7: Set Up the Equation for Gas Y
Using the ideal gas law for gas Y:
\[ P_Y V = n_Y R T \]
Substituting \( n_Y \):
\[ 0.5 \, \text{atm} \cdot V = \left(\frac{3}{M_Y}\right) R (298) \]

### Step 8: Relate the Two Equations
Now we have two equations:
1. For gas X:
\[ 1 \cdot V = \left(\frac{2}{M_X}\right) R (298) \]
2. For gas Y:
\[ 0.5 \cdot V = \left(\frac{3}{M_Y}\right) R (298) \]

### Step 9: Eliminate V and R
Dividing the two equations:
\[ \frac{1}{0.5} = \frac{\left(\frac{2}{M_X}\right)}{\left(\frac{3}{M_Y}\right)} \]
This simplifies to:
\[ 2 = \frac{2 M_Y}{3 M_X} \]

### Step 10: Rearranging to Find the Mass Ratio
Rearranging gives:
\[ 2 M_X = \frac{2}{3} M_Y \]
Thus:
\[ \frac{M_X}{M_Y} = \frac{1}{3} \]
Therefore:
\[ \frac{M_X}{M_Y} = \frac{3}{1} \]

### Final Result
The molecular mass ratio of \( M_X \) to \( M_Y \) is:
\[ \frac{M_X}{M_Y} = 3 \]