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This problem is asking to calculate the total volume of a gas mixture and the percentage volume contribution of neon, given the mass ratio of helium, argon, and neon in the mixture.

### Key information:
- The mass ratio of Helium (He), Argon (Ar), and Neon (Ne) is given as \(2:4:4\).
- The total mass of the gas mixture is \(100 \, \text{g}\).
- We need to find the volume of the gas mixture at standard conditions (i.e., 1 atm and 0°C) and the percentage of the volume contribution of Neon.

### Step 1: Assign masses based on the ratio
Let the masses of helium, argon, and neon be proportional to the given ratio \(2:4:4\). Since the total mass is \(100 \, \text{g}\), we distribute it as follows:
- \( \text{Mass of Helium (He)} = \frac{2}{10} \times 100 = 20 \, \text{g} \)
- \( \text{Mass of Argon (Ar)} = \frac{4}{10} \times 100 = 40 \, \text{g} \)
- \( \text{Mass of Neon (Ne)} = \frac{4}{10} \times 100 = 40 \, \text{g} \)

### Step 2: Calculate the number of moles of each gas
The number of moles \(n\) of a gas is given by:
\[
n = \frac{\text{mass}}{\text{molar mass}}
\]
The molar masses (in g/mol) of the gases are approximately:
- Helium (He): \(4 \, \text{g/mol}\)
- Argon (Ar): \(40 \, \text{g/mol}\)
- Neon (Ne): \(20 \, \text{g/mol}\)

Now, we calculate the moles of each gas:
- Moles of He: \( n_{\text{He}} = \frac{20}{4} = 5 \, \text{mol} \)
- Moles of Ar: \( n_{\text{Ar}} = \frac{40}{40} = 1 \, \text{mol} \)
- Moles of Ne: \( n_{\text{Ne}} = \frac{40}{20} = 2 \, \text{mol} \)

### Step 3: Calculate the total volume of the gas mixture
At standard conditions (1 atm and 0°C), 1 mole of any ideal gas occupies \(22.4 \, \text{L}\). Thus, the total volume \(V_{\text{total}}\) is:
\[
V_{\text{total}} = (n_{\text{He}} + n_{\text{Ar}} + n_{\text{Ne}}) \times 22.4
\]
\[
V_{\text{total}} = (5 + 1 + 2) \times 22.4 = 8 \times 22.4 = 179.2 \, \text{L}
\]

### Step 4: Calculate the percentage volume contribution of Neon
The volume contribution of Neon is proportional to the number of moles of Neon. The percentage volume contribution of Neon is:
\[
\% \, \text{Ne} = \frac{n_{\text{Ne}}}{n_{\text{total}}} \times 100 = \frac{2}{8} \times 100 = 25\%
\]

### Final Answers:
1. The total volume of the gas mixture is **179.2 liters**.
2. The volume contribution of Neon is **25%**.

Would you like further details on any part of the process, or do you have any questions?

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Here are some related questions:
1. How do different molar masses affect the calculation of gas volumes in mixtures?
2. What is the significance of standard conditions (STP) in gas law calculations?
3. How would the results change if the gases were at non-standard conditions?
4. Can you explain how the ideal gas law applies to this situation?
5. How does Dalton's Law of Partial Pressures relate to gas mixtures?

**Tip:** When dealing with gas mixtures, always remember that volume ratios for ideal gases at the same conditions of temperature and pressure are directly proportional to the number of moles.

Massa nisbatlari 2:4:4 bo'lgan geliy, argon va neondan iborat 100 g gazlar aralashmasining hajmini (l, n.sh.) va undagi neoning hajmiy ulushini (%) hisoblang.

Learn how to calculate the total volume of a gas mixture of helium, argon, and neon, and determine the percentage of neon in the mixture. This problem involves key concepts of gas laws and proportions, suitable for students in Grades 10-12.

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