A gaseous mixture containing 0.35g of `N_(2)` and 5600 ml of `O_(2)` at STP is kept in a 5 litres flask at 300K. The total pressure of the gaseous mixture is

Options

A. 1.293atm
B. 1.2315atm
C. 12.315atm
D. 0.616atm

Answer

To find the total pressure of the gaseous mixture, we will follow these steps:

### Step 1: Calculate the number of moles of Nitrogen (N₂)
Given:
- Mass of N₂ = 0.35 g
- Molar mass of N₂ = 28 g/mol

Using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
\[ \text{Moles of N₂} = \frac{0.35 \, \text{g}}{28 \, \text{g/mol}} = 0.0125 \, \text{mol} \]

### Step 2: Calculate the number of moles of Oxygen (O₂)
Given:
- Volume of O₂ at STP = 5600 mL = 5.6 L
- Molar volume of a gas at STP = 22.4 L/mol

Using the formula:
\[ \text{Number of moles} = \frac{\text{Volume at STP}}{\text{Molar volume}} \]
\[ \text{Moles of O₂} = \frac{5.6 \, \text{L}}{22.4 \, \text{L/mol}} = 0.25 \, \text{mol} \]

### Step 3: Calculate the total number of moles in the mixture
\[ \text{Total moles} (n) = \text{Moles of N₂} + \text{Moles of O₂} \]
\[ n = 0.0125 \, \text{mol} + 0.25 \, \text{mol} = 0.2625 \, \text{mol} \]

### Step 4: Use the Ideal Gas Law to find the total pressure
The Ideal Gas Law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure (atm)
- \( V \) = volume (L)
- \( n \) = number of moles (mol)
- \( R \) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature (K)

Given:
- Volume \( V = 5 \, \text{L} \)
- Temperature \( T = 300 \, \text{K} \)

Rearranging the Ideal Gas Law to solve for \( P \):
\[ P = \frac{nRT}{V} \]
Substituting the values:
\[ P = \frac{0.2625 \, \text{mol} \times 0.0821 \, \text{L·atm/(K·mol)} \times 300 \, \text{K}}{5 \, \text{L}} \]
Calculating:
\[ P = \frac{0.2625 \times 0.0821 \times 300}{5} \]
\[ P = \frac{6.467325}{5} = 1.293465 \, \text{atm} \]

### Final Answer
The total pressure of the gaseous mixture is approximately **1.293 atm**.

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