Which of the following options does not represent concentration of semi-molal aqueous solution of NaOH having `d_("solution")` = 1.02 g/ml?

Options

A. Molarity = `(1)/(2)` M
B. `X_(NaOH) = (9)/(1009)`
C. `% (w)/(w) = 10 %`
D. `% (w)/(v) = 2%`

Answer

To determine which option does not represent the concentration of a semi-molal aqueous solution of NaOH with a density of 1.02 g/ml, we will analyze the given options step by step.

### Step 1: Understanding Semi-Molal Solution
A semi-molal solution means that there are 0.5 moles of solute (NaOH) per 1 kg of solvent (water).

### Step 2: Calculate the Mass of NaOH
1 mole of NaOH has a molar mass of approximately 40 g. Therefore, 0.5 moles of NaOH will have:
\[ \text{Mass of NaOH} = 0.5 \, \text{moles} \times 40 \, \text{g/mole} = 20 \, \text{g} \]

### Step 3: Calculate the Total Mass of the Solution
The total mass of the solution is the mass of the solvent plus the mass of the solute:
\[ \text{Mass of solution} = \text{Mass of water} + \text{Mass of NaOH} = 1000 \, \text{g} + 20 \, \text{g} = 1020 \, \text{g} \]

### Step 4: Calculate the Volume of the Solution
Using the density of the solution, we can calculate the volume:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \implies \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{1020 \, \text{g}}{1.02 \, \text{g/ml}} = 1000 \, \text{ml} \]

### Step 5: Calculate Molarity
Molarity (M) is defined as the number of moles of solute per liter of solution. Since we have 0.5 moles of NaOH in 1 liter (1000 ml) of solution:
\[ \text{Molarity} = \frac{0.5 \, \text{moles}}{1 \, \text{L}} = 0.5 \, \text{M} \]

### Step 6: Calculate Mole Fraction of NaOH
The mole fraction (X) of NaOH is calculated as:
\[ X_{\text{NaOH}} = \frac{\text{Moles of NaOH}}{\text{Moles of NaOH} + \text{Moles of water}} \]
The moles of water can be calculated as:
\[ \text{Moles of water} = \frac{1000 \, \text{g}}{18 \, \text{g/mole}} \approx 55.56 \, \text{moles} \]
Thus, the mole fraction is:
\[ X_{\text{NaOH}} = \frac{0.5}{0.5 + 55.56} \approx \frac{0.5}{56.06} \approx 0.0089 \]

### Step 7: Calculate Weight by Weight Percentage (W/W%)
The W/W% is calculated as:
\[ \text{W/W\%} = \frac{\text{Mass of NaOH}}{\text{Total mass of solution}} \times 100 = \frac{20 \, \text{g}}{1020 \, \text{g}} \times 100 \approx 1.96\% \]

### Step 8: Calculate Weight by Volume Percentage (W/V%)
The W/V% is calculated as:
\[ \text{W/V\%} = \frac{\text{Mass of NaOH}}{\text{Volume of solution}} \times 100 = \frac{20 \, \text{g}}{1000 \, \text{ml}} \times 100 = 2\% \]

### Conclusion
Now we can summarize the results:
- Molarity: 0.5 M (Correct)
- Mole Fraction: Approximately 0.0089 (Correct)
- W/W%: Approximately 1.96% (Incorrectly stated as 10%)
- W/V%: 2% (Correct)

Thus, the option that does not represent the concentration correctly is the W/W% which is stated as 10%.

### Final Answer
**The option that does not represent the concentration correctly is the W/W% which is 10%.**

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