Question

A container contains 4 g $H_2$, 4 g $He$ and certain amount of 'Ne' at a certain temperature. What is the mass of 'Ne' required so that the partial pressure exerted by 'Ne' is equal to the partial pressure of He?

• A. 4 g
• B. 8 g
• C. 10 g
• D. 20 g

Hint:

In identical physical conditions ($V$ and $T$), gases act entirely independently. "Equal partial pressures" means "Equal number of moles" or "Equal number of molecules", regardless of how heavy those individual molecules are.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A gas mixture contains multiple gases. We are told the partial pressure of Neon ($Ne$) must equal the partial pressure of Helium ($He$). We must calculate the mass of Neon required to achieve this.

Step 2: Key Formula or Approach:
According to Dalton's Law of Partial Pressures and the Ideal Gas Law ($PV=nRT$), if multiple gases are in the same container at the same volume and temperature, their partial pressures are directly proportional to their number of moles.
$$P_A \propto n_A$$
Therefore, if $P_{Ne} = P_{He}$, then it must strictly follow that $n_{Ne} = n_{He}$.

Step 3: Detailed Explanation:
1. First, calculate the number of moles of Helium present.
Mass of $He$ = 4 g.
Molar mass of $He$ = 4 g/mol.
$$n_{He} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{4}{4} = 1 \text{ mole}$$
2. Since partial pressures must be equal, the moles must be equal.
$$n_{Ne} = n_{He} = 1 \text{ mole}$$
3. Now, convert this required 1 mole of Neon back into mass.
Molar mass of Neon ($Ne$) $\approx$ 20 g/mol.
$$\text{Mass of } Ne = n_{Ne} \times \text{Molar Mass}$$
$$\text{Mass of } Ne = 1 \text{ mol} \times 20 \text{ g/mol} = 20 \text{ g}$$
(Note: The 4g of $H_2$ is extraneous information designed to distract you).

Step 4: Final Answer:
The required mass of Neon is 20 g, matching option (d).