Three statements are given about mole fraction
(i) Mole fraction of a solute + mole fraction of solvent = 1
(ii) Equal weights of Helium and methane are present in a gaseous mixture. The mole fraction of He is 4/5
(iii) The mole fraction of water in the aqueous solution of NaOH is 0.8. The molality of the solution is nearly 14 moles `kg^(-1)`

Options

A. i and ii are correct
B. ii and iii are correct
C. i and iii are correct
D. all are correct

Answer

To solve the question regarding the three statements about mole fraction, we will analyze each statement step by step.

### Step 1: Analyze Statement 1
**Statement:** Mole fraction of a solute + mole fraction of solvent = 1.

**Solution:**
- The mole fraction (X) of a component in a solution is defined as the ratio of the number of moles of that component to the total number of moles of all components in the solution.
- For a solution containing a solute and a solvent:
\[ X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \]
\[ X_{\text{solvent}} = \frac{n_{\text{solvent}}}{n_{\text{solute}} + n_{\text{solvent}}} \]
- Adding these two equations:
\[ X_{\text{solute}} + X_{\text{solvent}} = \frac{n_{\text{solute}} + n_{\text{solvent}}}{n_{\text{solute}} + n_{\text{solvent}}} = 1 \]
- Therefore, **Statement 1 is correct.**

### Step 2: Analyze Statement 2
**Statement:** Equal weights of Helium and Methane are present in a gaseous mixture. The mole fraction of He is 4/5.

**Solution:**
- Let the weight of Helium (He) and Methane (CH₄) be \(x\) grams each.
- Moles of Helium:
\[ n_{\text{He}} = \frac{x}{4} \quad (\text{Molar mass of He} = 4 \text{ g/mol}) \]
- Moles of Methane:
\[ n_{\text{CH}_4} = \frac{x}{16} \quad (\text{Molar mass of CH₄} = 16 \text{ g/mol}) \]
- Total moles in the mixture:
\[ n_{\text{total}} = n_{\text{He}} + n_{\text{CH}_4} = \frac{x}{4} + \frac{x}{16} \]
To combine these, find a common denominator:
\[ n_{\text{total}} = \frac{4x}{16} + \frac{x}{16} = \frac{5x}{16} \]
- Mole fraction of Helium:
\[ X_{\text{He}} = \frac{n_{\text{He}}}{n_{\text{total}}} = \frac{\frac{x}{4}}{\frac{5x}{16}} = \frac{16}{20} = \frac{4}{5} \]
- Therefore, **Statement 2 is correct.**

### Step 3: Analyze Statement 3
**Statement:** The mole fraction of water in the aqueous solution of NaOH is 0.8. The molality of the solution is nearly 14 moles kg⁻¹.

**Solution:**
- Given \(X_{\text{water}} = 0.8\), we can find the mole fraction of NaOH:
\[ X_{\text{NaOH}} = 1 - X_{\text{water}} = 0.2 \]
- Let the number of moles of water be \(x\). Then, the number of moles of NaOH is:
\[ n_{\text{NaOH}} = \frac{0.2}{0.8}x = \frac{x}{4} \]
- The weight of water in grams:
\[ \text{Weight of water} = 18x \text{ grams} \quad (\text{Molar mass of water} = 18 g/mol) \]
- Convert to kg:
\[ \text{Weight of water in kg} = \frac{18x}{1000} \]
- Molality (m) is defined as:
\[ m = \frac{n_{\text{solute}}}{\text{weight of solvent in kg}} = \frac{\frac{x}{4}}{\frac{18x}{1000}} = \frac{1000}{72} \approx 13.89 \text{ moles kg}^{-1} \]
- This rounds to approximately 14 moles kg⁻¹, hence **Statement 3 is correct.**

### Conclusion
All three statements are correct.

### Final Answer
**All statements are correct.**