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Preparation of Benzene and Homologues

224 mL of SO_...

Question

$224 mL$ 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 $x \times 10^{- 2} mm of Hg$ , the value of $x$ is _____

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Solution

Step 1: Given data:
Volume of ${SO}_{2} (g)$ is $V_{1} = 224 mL$
Volume of $NaOH$ is $V_{2} = 100 mL$
Molarity of $NaOH$ is $M_{2} = 0.1 M$
Weight of $H_{2} O$ is $w_{(H_{2} O)} = 36 g$
Temperature is $T = 298 K$
Pressure is $P = 1 atm$
Step 2: Write the balanced equation of ${SO}_{2}$ and $NaOH$ :
${SO}_{2} + 2 NaOH \to {Na}_{2} {SO}_{3} + H_{2} O$
Here, NaOH acts as a Limiting reagent.
Step 3: Find the number of moles of $NaOH$ , $H_{2} O$ and ${Na}_{2} {SO}_{3}$ :
Number of moles of $NaOH$ is
$n_{(NaOH)} = Molarity of NaOH \times Volume of NaOH \Rightarrow n_{(NaOH)} = 0.1 M \times 0.1 L \Rightarrow n_{(NaOH)} = 0.0.1 moles$
2. Number of moles of ${Na}_{2} {SO}_{3}$ is
$2 moles of NaOH \to 1 mole of {Na}_{2} {SO}_{3} \Rightarrow 0.01 moles of NaOH \to ? \Rightarrow n_{({Na}_{2} {SO}_{3})} = \frac{0.01 \times 1}{2} \Rightarrow n_{({Na}_{2} {SO}_{3})} = 0.05 moles$
3. Number of moles of $H_{2} O$ is
$n_{(H_{2} O)} = \frac{w_{(H_{2} O)}}{M . W ._{(H_{2} O)}} \Rightarrow n_{(H_{2} O)} = \frac{36 g}{18 g {mol}^{- 1}} \Rightarrow n_{(H_{2} O)} = 2 moles$
Step 4: Find the van't Hoff factor $(i)$ :
${Na}_{2} {SO}_{3} \to 2 {Na}^{+} + {SO}_{3}^{- 2}$
Total number of ions formed is $i = 3$
Step 5:Find the Relative Lowering of Vapour Pressure $(RLVP)$ :
$\frac{∆ P_{A}}{P_{A}^{0}} = i χ_{B} \Rightarrow \frac{∆ P_{A}}{P_{A}^{0}} = \frac{{in}_{A}}{{in}_{A} + n_{B}} [∵ {in}_{A} \approx 0] \Rightarrow \frac{∆ P_{A}}{24 mm Hg} = i \times \frac{n_{A}}{n_{B}} [∵ given P_{(H_{2} O)} = 24 mm of Hg] \Rightarrow \frac{∆ P_{A}}{24} = 3 \times \frac{0.005}{2} \Rightarrow ∆ P_{A} = 0.18 \Rightarrow ∆ P_{A} = 18 \times 10^{- 2} mm of Hg$
Hence, the lowering of vapour pressure of the solution is $∆ P_{A} = 18 \times 10^{- 2} mm of Hg$ .
The value of $x = 18$ .

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