15g urea and 20g NaOH dissolved in water. Total mass in solution is 250g. Mole fraction of NaOH in the mixture.

Options

A. 0.039
B. 0.62
C. 0.5
D. 0.4

Answer

To find the mole fraction of NaOH in the solution, we will follow these steps:

### Step 1: Calculate the number of moles of urea.
- **Given:** Mass of urea = 15 g
- **Molecular weight of urea (CH₃CONH₂):**
- Carbon (C) = 12 g/mol (1 atom)
- Hydrogen (H) = 1 g/mol (6 atoms)
- Nitrogen (N) = 14 g/mol (1 atom)
- Oxygen (O) = 16 g/mol (1 atom)

Molecular weight = 12 + (3 × 1) + 12 + 16 + 14 + (2 × 1) = 60 g/mol

- **Number of moles of urea:**
\[ \text{Moles of urea} = \frac{\text{mass}}{\text{molecular weight}} = \frac{15 \, \text{g}}{60 \, \text{g/mol}} = 0.25 \, \text{moles} \]

### Step 2: Calculate the number of moles of NaOH.
- **Given:** Mass of NaOH = 20 g
- **Molecular weight of NaOH:**
- Sodium (Na) = 23 g/mol (1 atom)
- Oxygen (O) = 16 g/mol (1 atom)
- Hydrogen (H) = 1 g/mol (1 atom)

Molecular weight = 23 + 16 + 1 = 40 g/mol

- **Number of moles of NaOH:**
\[ \text{Moles of NaOH} = \frac{20 \, \text{g}}{40 \, \text{g/mol}} = 0.5 \, \text{moles} \]

### Step 3: Calculate the mass of water in the solution.
- **Total mass of the solution = 250 g**
- **Mass of water:**
\[ \text{Mass of water} = \text{Total mass} - (\text{mass of urea} + \text{mass of NaOH}) = 250 \, \text{g} - (15 \, \text{g} + 20 \, \text{g}) = 215 \, \text{g} \]

### Step 4: Calculate the number of moles of water.
- **Molecular weight of water (H₂O):**
- Hydrogen (H) = 1 g/mol (2 atoms)
- Oxygen (O) = 16 g/mol (1 atom)

Molecular weight = (2 × 1) + 16 = 18 g/mol

- **Number of moles of water:**
\[ \text{Moles of water} = \frac{215 \, \text{g}}{18 \, \text{g/mol}} \approx 11.94 \, \text{moles} \]

### Step 5: Calculate the total number of moles in the solution.
\[ \text{Total moles} = \text{Moles of urea} + \text{Moles of NaOH} + \text{Moles of water} = 0.25 + 0.5 + 11.94 \approx 12.69 \, \text{moles} \]

### Step 6: Calculate the mole fraction of NaOH.
\[ \text{Mole fraction of NaOH} = \frac{\text{Moles of NaOH}}{\text{Total moles}} = \frac{0.5}{12.69} \approx 0.0394 \]

### Final Answer:
The mole fraction of NaOH in the mixture is approximately **0.039**.

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