8g of NaOH is dissolved in 18g of `H_(2)O`. Mole fraction of NaOH in solution and molality (in mol `kg^(-1)`) of the solutions respectively are:

Options

A. `0.1,11.11`
B. `0.167,11.11`
C. `0.2,22.20`
D. `0.167,22.20`

Explanation

To solve the problem, we need to find the mole fraction of NaOH in the solution and the molality of the solution. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the number of moles of NaOH
1. **Given mass of NaOH** = 8 g
2. **Molar mass of NaOH** = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol
3. **Number of moles of NaOH** = Mass / Molar mass = 8 g / 40 g/mol = 0.2 moles

### Step 2: Calculate the number of moles of water (H₂O)
1. **Given mass of water** = 18 g
2. **Molar mass of H₂O** = 2 (H) + 16 (O) = 18 g/mol
3. **Number of moles of H₂O** = Mass / Molar mass = 18 g / 18 g/mol = 1 mole

### Step 3: Calculate the total number of moles in the solution
1. **Total moles** = Moles of NaOH + Moles of H₂O = 0.2 moles + 1 mole = 1.2 moles

### Step 4: Calculate the mole fraction of NaOH
1. **Mole fraction of NaOH (χ)** = Moles of NaOH / Total moles = 0.2 moles / 1.2 moles = 0.167

### Step 5: Calculate the molality of the solution
1. **Mass of solvent (water)** = 18 g = 0.018 kg
2. **Molality (m)** = Moles of solute / Mass of solvent in kg = 0.2 moles / 0.018 kg
3. **Molality** = 0.2 / 0.018 = 11.11 mol/kg

### Final Results:
- Mole fraction of NaOH = 0.167
- Molality of the solution = 11.11 mol/kg

### Summary:
The mole fraction of NaOH in the solution is **0.167**, and the molality of the solution is **11.11 mol/kg**.

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