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Step-by-Step Solution
Step 1: Calculate moles of SO2
Given volume of SO2 at NTP conditions (1 atm and 298 K, assuming close to standard conditions) is 224 mL. The molar volume at NTP is approximately 22.4 L (or 22400 mL) per mole. Hence,
\text{Moles of } SO_2 = \dfrac{224 \text{ mL}}{22400 \text{ mL/mol}} = 0.01 \text{ moles}
Step 2: Calculate moles of NaOH
Molarity (M) = 0.1 mol/L
Volume = 100 mL = 0.1 L
\text{Moles of NaOH} = \text{Molarity} \times \text{Volume in litres} = 0.1 \times 0.1 = 0.01 \text{ moles}
Step 3: Identify the limiting reagent and form the product
The balanced reaction for the absorption of SO2 by NaOH is:
SO_2 + 2 \, NaOH \longrightarrow Na_2SO_3 + H_2O
For every 1 mole of SO2, 2 moles of NaOH are required.
Here, we have 0.01 moles of both SO2 and NaOH, but since the reaction requires twice as many moles of NaOH, NaOH becomes the limiting reagent.
Thus, if 2 moles of NaOH produce 1 mole of Na2SO3, then
0.01 \text{ moles NaOH} \longrightarrow \dfrac{0.01}{2} = 0.005 \text{ moles Na}_2\text{SO}_3
Step 4: Determine van’t Hoff factor (i)
Sodium sulphite ( Na_2SO_3 ) dissociates in water as:
Na_2SO_3 \longrightarrow 2 Na^+ + SO_3^{2-}
Thus, 1 mole of Na_2SO_3 forms 3 total ions. Therefore, the van’t Hoff factor i = 3 .
Step 5: Calculate moles of water used as solvent
36 g of water has mole count:
\text{Moles of water} = \dfrac{36 \text{ g}}{18 \text{ g/mol}} = 2 \text{ moles}
Step 6: Use the formula for relative lowering of vapour pressure
The relative lowering of vapour pressure for a dilute solution can be written as:
\dfrac{P_{{H_2}O}^o - P_S}{P_{{H_2}O}^o} \;=\; \dfrac{i \, n_{\text{solute}}}{n_{\text{solvent}} + i \, n_{\text{solute}}}
Since n_{\text{solute}} \ll n_{\text{solvent}} for a dilute solution, we approximate:
\dfrac{P_{{H_2}O}^o - P_S}{P_{{H_2}O}^o} \; \approx \; \dfrac{i \, n_{\text{solute}}}{n_{\text{solvent}}}
Here,
P_{{H_2}O}^o = 24 \text{ mmHg (vapour pressure of pure water)} ,
n_{\text{solute}} = 0.005 \text{ moles of } Na_2SO_3 ,
i = 3 ,
n_{\text{solvent}} = 2 \text{ moles of } H_2O .
So,
\dfrac{24 - P_S}{24} = \dfrac{3 \times 0.005}{2} = \dfrac{0.015}{2} = 0.0075
24 - P_S = 24 \times 0.0075 = 0.18
P_S = 24 - 0.18 = 23.82 \text{ mmHg}
Thus, the lowering of vapour pressure \Delta P is:
\Delta P = 24 - 23.82 = 0.18 \text{ mmHg}
Step 7: Express the lowering of vapour pressure in the desired form
The problem states the lowering in vapour pressure is x \times 10^{-2} \text{ mmHg} . We have:
0.18 = 18 \times 10^{-2}
Hence, x = 18 .
Final Answer
x = 18
