Aqueous solution of NaOH is marked 10% `(w//w)`. The density of the solution is 1.070 g `cm^(-3)`. Calculate (i) molarity and (ii) molality of NaOH.

Explanation

To solve the problem, we need to calculate both the molarity and molality of the NaOH solution. Let's break it down step by step.

### Given Data:
- Weight percent (w/w) of NaOH = 10%
- Density of the solution = 1.070 g/cm³

### Step 1: Calculate the mass of NaOH in 100 g of solution
Since the solution is 10% w/w, this means that in 100 g of the solution, there are 10 g of NaOH.

### Step 2: Calculate the molar mass of NaOH
The molar mass of NaOH can be calculated as follows:
- Na: 23 g/mol
- O: 16 g/mol
- H: 1 g/mol

So, the molar mass of NaOH = 23 + 16 + 1 = 40 g/mol.

### Step 3: Calculate the number of moles of NaOH
Using the formula:
\[ \text{Number of moles} (n) = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ n = \frac{10 \text{ g}}{40 \text{ g/mol}} = 0.25 \text{ moles} \]

### Step 4: Calculate the volume of the solution
Using the density formula:
\[ \text{Density} = \frac{\text{mass}}{\text{volume}} \]
Rearranging gives us:
\[ \text{Volume} = \frac{\text{mass}}{\text{density}} \]
The mass of the solution is 100 g, and the density is 1.070 g/cm³:
\[ \text{Volume} = \frac{100 \text{ g}}{1.070 \text{ g/cm}^3} \approx 93.46 \text{ cm}^3 \]
Converting cm³ to liters:
\[ \text{Volume} = 93.46 \text{ cm}^3 \times \frac{1 \text{ L}}{1000 \text{ cm}^3} = 0.09346 \text{ L} \]

### Step 5: Calculate the molarity of NaOH
Using the formula for molarity (M):
\[ M = \frac{n}{V} \]
Substituting the values:
\[ M = \frac{0.25 \text{ moles}}{0.09346 \text{ L}} \approx 2.676 \text{ mol/L} \]

### Step 6: Calculate the mass of the solvent (water)
The mass of the solvent can be calculated by subtracting the mass of NaOH from the total mass of the solution:
\[ \text{Mass of solvent} = 100 \text{ g} - 10 \text{ g} = 90 \text{ g} = 0.090 \text{ kg} \]

### Step 7: Calculate the molality of NaOH
Using the formula for molality (m):
\[ m = \frac{n}{\text{mass of solvent (kg)}} \]
Substituting the values:
\[ m = \frac{0.25 \text{ moles}}{0.090 \text{ kg}} \approx 2.778 \text{ mol/kg} \]

### Final Answers:
(i) Molarity of NaOH = 2.676 mol/L
(ii) Molality of NaOH = 2.778 mol/kg