What will be the value of molality for an aqueous solution of 10% w/W NaOH?

Options

A. 0.01 M
B. 0.617 M
C. 0.668 M
D. 1.623 M

Answer

To find the molality of a 10% w/w NaOH solution, we can follow these steps:

### Step 1: Understand the given data
We have a 10% w/w solution of NaOH. This means that in 100 grams of the solution, there are 10 grams of NaOH.

### Step 2: Calculate the weight of the solvent
The total weight of the solution is 100 grams, and since we have 10 grams of NaOH, the weight of the solvent (water) can be calculated as:
\[ \text{Weight of solvent} = \text{Weight of solution} - \text{Weight of solute} = 100 \, \text{g} - 10 \, \text{g} = 90 \, \text{g} \]
Now, convert this weight into kilograms:
\[ \text{Weight of solvent in kg} = \frac{90 \, \text{g}}{1000} = 0.09 \, \text{kg} \]

### Step 3: Calculate the number of moles of NaOH
To find the number of moles of NaOH, we use the formula:
\[ \text{Number of moles} = \frac{\text{Weight of NaOH}}{\text{Molecular weight of NaOH}} \]
The molecular weight of NaOH is calculated as follows:
- Sodium (Na) = 23 g/mol
- Oxygen (O) = 16 g/mol
- Hydrogen (H) = 1 g/mol

Thus, the molecular weight of NaOH is:
\[ \text{Molecular weight of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol} \]
Now, substituting the values:
\[ \text{Number of moles of NaOH} = \frac{10 \, \text{g}}{40 \, \text{g/mol}} = 0.25 \, \text{moles} \]

### Step 4: Calculate the molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} = \frac{\text{Number of moles of NaOH}}{\text{Weight of solvent in kg}} = \frac{0.25 \, \text{moles}}{0.09 \, \text{kg}} \approx 2.77 \, \text{mol/kg} \]

### Final Answer
The molality of the 10% w/w NaOH solution is approximately **2.77 mol/kg**.

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