224 mL of `SO_(2(g))` at 298 K and 1 atm is passed through 100 mL of 0.1 M NaOH solution. The non-volatile solute produced is dissolved in 36 g of water. The lowering of vapour pressure of solution (assuming the solution is dilute)
`(P_((H_2 O)) = 24 "mm of Hg) is" xx 10^(-2)` mm of Hg, the value of x is _________. (Integer answer)

Answer

To solve the problem, we need to calculate the lowering of vapor pressure of the solution formed when `SO2` reacts with `NaOH`. Let's break down the solution step by step.

### Step 1: Calculate the number of moles of `SO2`
Given:
- Volume of `SO2` = 224 mL
- Temperature = 298 K
- Pressure = 1 atm

Using the ideal gas law, we can calculate the number of moles of `SO2` at STP (Standard Temperature and Pressure). At STP, 1 mole of gas occupies 22.4 L (or 22400 mL).

\[ \text{Number of moles of } SO_2 = \frac{224 \text{ mL}}{22400 \text{ mL/mol}} = 0.01 \text{ moles} \]

### Step 2: Calculate the number of moles of `NaOH`
Given:
- Volume of `NaOH` solution = 100 mL
- Molarity of `NaOH` = 0.1 M

Using the formula for moles:

\[ \text{Number of moles of } NaOH = \text{Molarity} \times \text{Volume (in L)} = 0.1 \text{ mol/L} \times 0.1 \text{ L} = 0.01 \text{ moles} \]

### Step 3: Determine the limiting reagent
The balanced reaction is:

\[ 2 NaOH + SO_2 \rightarrow Na_2SO_3 + H_2O \]

From the balanced equation, 2 moles of `NaOH` react with 1 mole of `SO2`.

- Moles of `NaOH` required for 0.01 moles of `SO2` = \(2 \times 0.01 = 0.02\) moles.
- Available moles of `NaOH` = 0.01 moles.

Since we have only 0.01 moles of `NaOH`, it is the limiting reagent.

### Step 4: Calculate moles of `Na2SO3` produced
From the stoichiometry of the reaction, 2 moles of `NaOH` produce 1 mole of `Na2SO3`. Thus:

\[ \text{Moles of } Na_2SO_3 = \frac{0.01 \text{ moles of } NaOH}{2} = 0.005 \text{ moles} \]

### Step 5: Calculate the number of moles of water
Given:
- Mass of water = 36 g
- Molar mass of water (H2O) = 18 g/mol

\[ \text{Number of moles of water} = \frac{36 \text{ g}}{18 \text{ g/mol}} = 2 \text{ moles} \]

### Step 6: Calculate the lowering of vapor pressure
Using the formula for relative lowering of vapor pressure:

\[ \frac{\Delta P}{P_0} = i \cdot \frac{n_{\text{solute}}}{n_{\text{solvent}}} \]

Where:
- \(i\) = van 't Hoff factor (for `Na2SO3`, it dissociates into 3 ions: 2 Na⁺ and SO3²⁻, so \(i = 3\))
- \(n_{\text{solute}} = 0.005\) moles of `Na2SO3`
- \(n_{\text{solvent}} = 2\) moles of water
- \(P_0\) (vapor pressure of pure water) = 24 mmHg

Substituting the values:

\[ \frac{\Delta P}{24} = 3 \cdot \frac{0.005}{2} \]

Calculating the right side:

\[ \frac{\Delta P}{24} = 3 \cdot 0.0025 = 0.0075 \]

Now, solving for \(\Delta P\):

\[ \Delta P = 0.0075 \times 24 = 0.18 \text{ mmHg} \]

### Step 7: Express the result in the required format
The problem states that the lowering of vapor pressure is \(xx \times 10^{-2}\) mmHg. We have:

\[ \Delta P = 0.18 \text{ mmHg} = 18 \times 10^{-2} \text{ mmHg} \]

Thus, the value of \(x\) is:

\[ \boxed{18} \]