Question

A neon-dioxygen mixture contains 64 g $O_2$ and 160 g $Ne$. If the total pressure is 25 bar, calculate the partial pressure of dioxygen.

• A. 5 bar
• B. 7.5 bar
• C. 10 bar
• D. 20 bar

Hint:

Dalton's Law of Partial Pressures states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases. The partial pressure of a gas is its mole fraction multiplied by the total pressure: $P_i = X_i \times P_{\text{Total$.

Answer 1

The Correct Option is B

Step 1: Calculate the number of moles for $O_2$ and $Ne$.\ \ Given mass of $O_2 = 64\text{ g}$\ Molar mass of $O_2 = 32\text{ g/mol}$\ Number of moles of $O_2$ ($n_{O_2}$):\ \[ n_{O_2} = \frac{\text{Mass of } O_2}{\text{Molar mass of } O_2} = \frac{64\text{ g{32\text{ g/mol = 2\text{ mol} \]\ Given mass of $Ne = 160\text{ g}$\ Molar mass of $Ne = 20\text{ g/mol}$\ Number of moles of $Ne$ ($n_{Ne}$):\ \[ n_{Ne} = \frac{\text{Mass of } Ne}{\text{Molar mass of } Ne} = \frac{160\text{ g{20\text{ g/mol = 8\text{ mol} \]\
Step 2: Calculate the total number of moles.\ Total number of moles ($n_{\text{Total$) in the mixture:\ \[ n_{\text{Total = n_{O_2} + n_{Ne} = 2\text{ mol} + 8\text{ mol} = 10\text{ mol} \]\
Step 3: Calculate the mole fraction of $O_2$.\ Mole fraction of $O_2$ ($X_{O_2}$):\ \[ X_{O_2} = \frac{n_{O_2{n_{\text{Total} = \frac{2\text{ mol{10\text{ mol = 0.2 \]\
Step 4: Calculate the partial pressure of $O_2$.\ Given total pressure ($P_{\text{Total$) = 25 bar\ Partial pressure of $O_2$ ($P_{O_2}$) can be calculated using Dalton's Law of Partial Pressures:\ \[ P_{O_2} = X_{O_2} \times P_{\text{Total = 0.2 \times 25\text{ bar} = 5\text{ bar} \]\ Therefore, the partial pressure of dioxygen is 5 bar.