To solve the question step by step, we need to calculate the mole fraction, molality, molarity, and normality of the NaOH solution given that there are 4 g of NaOH in 0.1 dm³ of solution.
### Step 1: Calculate the weight of the solution
The volume of the solution is given as 0.1 dm³, which is equal to 100 mL. The density of the solution is given as 1.038 g/mL.
**Weight of the solution = Volume × Density**
\[
\text{Weight of the solution} = 100 \, \text{mL} \times 1.038 \, \text{g/mL} = 103.8 \, \text{g}
\]
### Step 2: Calculate the mass of water
To find the mass of water in the solution, we subtract the mass of NaOH from the total mass of the solution.
**Mass of water = Mass of solution - Mass of NaOH**
\[
\text{Mass of water} = 103.8 \, \text{g} - 4 \, \text{g} = 99.8 \, \text{g}
\]
### Step 3: Calculate the mole fraction of NaOH
The mole fraction of NaOH is calculated using the formula:
\[
\text{Mole fraction of NaOH} = \frac{\text{Number of moles of NaOH}}{\text{Total moles of NaOH and water}}
\]
**Calculate moles of NaOH:**
\[
\text{Moles of NaOH} = \frac{\text{Mass of NaOH}}{\text{Molar mass of NaOH}} = \frac{4 \, \text{g}}{40 \, \text{g/mol}} = 0.1 \, \text{mol}
\]
**Calculate moles of water:**
\[
\text{Moles of water} = \frac{\text{Mass of water}}{\text{Molar mass of water}} = \frac{99.8 \, \text{g}}{18 \, \text{g/mol}} \approx 5.55 \, \text{mol}
\]
**Total moles:**
\[
\text{Total moles} = 0.1 + 5.55 = 5.65 \, \text{mol}
\]
**Mole fraction of NaOH:**
\[
\text{Mole fraction of NaOH} = \frac{0.1}{5.65} \approx 0.0177
\]
### Step 4: Calculate the molality of NaOH solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent.
\[
\text{Molality} = \frac{\text{Moles of NaOH}}{\text{Mass of solvent in kg}} = \frac{0.1 \, \text{mol}}{0.0998 \, \text{kg}} \approx 1.002 \, \text{mol/kg}
\]
### Step 5: Calculate the molarity of NaOH solution
Molarity (M) is defined as the number of moles of solute per liter of solution.
\[
\text{Molarity} = \frac{\text{Moles of NaOH}}{\text{Volume of solution in L}} = \frac{0.1 \, \text{mol}}{0.1 \, \text{L}} = 1 \, \text{M}
\]
### Step 6: Calculate the normality of NaOH solution
Normality (N) is defined as the number of equivalents of solute per liter of solution. For NaOH, the n-factor is 1.
\[
\text{Normality} = \text{Molarity} \times n = 1 \, \text{M} \times 1 = 1 \, \text{N}
\]
### Summary of Results:
(a) Mole fraction of NaOH ≈ 0.0177
(b) Molality of NaOH solution ≈ 1.002 mol/kg
(c) Molarity of NaOH solution = 1 M
(d) Normality of NaOH solution = 1 N
