Choose the correct option for the total pressure (in atm.) in a mixture of 4 g O₂ and 2 g H₂ confined in a total volume of one litre at 0°C is : [Given R=0.082 L atm mol⁻¹K⁻¹, T=273 K]

The correct option is 25.18.

To find the total pressure exerted by the mixture of gases, we can use Dalton's Law of Partial Pressures and the ideal gas equation. Dalton's law states that the total pressure of a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases, which can also be calculated using the total number of moles of the mixture in the ideal gas equation:

$P=\frac{n_{\text{total}}RT}{V}$

Step 1: Calculate the number of moles of each gas
The molar mass of Oxygen ($O_2$) is 32 g/mol, and the molar mass of Hydrogen ($H_2$) is 2 g/mol.

For Oxygen ($O_2$):
$n_{O_2}=\frac{4\text{g}}{32\text{g/mol}}=0.125\text{mol}$

For Hydrogen ($H_2$):
$n_{H_2}=\frac{2\text{g}}{2\text{g/mol}}=1.0\text{mol}$

Step 2: Calculate the total number of moles in the mixture
$n_{\text{total}}=n_{O_2}+n_{H_2}$
$n_{\text{total}}=0.125\text{mol}+1.0\text{mol}=1.125\text{mol}$

Step 3: Calculate the total pressure using the ideal gas law
Given parameters:
Volume (V) = 1 L
Temperature (T) = 273 K
Gas constant (R) = 0.082 L atm mol^-1 K^-1

Substitute these values into the equation:
$P=\frac{1.125\text{mol}\times 0.082\text{L atm}{\text{mol}}^{-1}{\text{K}}^{-1}\times 273\text{K}}{1\text{L}}$

Perform the multiplication:
$P=1.125\times 22.386\text{atm}=25.18425\text{atm}$

Rounding to two decimal places, we get:
$P\approx 25.18\text{atm}$

Therefore, the total pressure in the vessel is 25.18 atm.