1.22 g of an organic acid is separately dissolved in 100 g of benzene ($K_b = 2.6 \text{ K kg mol}^{-1}$) and 100 g of acetone ($K_b = 1.7 \text{ K kg mol}^{-1}$). The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by $0.17^\circ\text{C}$. The increase in boiling point of solution in benzene in $^\circ\text{C}$ is $x \times 10^{-2}$. The value of $x$ is ________. (Nearest integer) [Atomic mass : C = 12.0, H = 1.0, O = 16.0]
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Dimerization decreases the number of particles by half, so the van't Hoff factor $i$ becomes 0.5. Consequently, for the same molality, the colligative effect is halved.
Step 1: Understanding the Concept:
The elevation in boiling point ($\Delta T_b$) is a colligative property given by $\Delta T_b = i \times K_b \times m$, where $i$ is the van't Hoff factor. The factor $i$ accounts for association or dissociation of the solute. Step 2: Key Formula or Approach:
1. $\Delta T_b = i K_b \frac{w_B \times 1000}{M_B \times w_A}$
2. For monomer in acetone, $i = 1$.
3. For dimer in benzene (assuming 100% dimerization), $i = 1 - \alpha/2 = 0.5$. Step 3: Detailed Explanation:
1. From the acetone solution, find the molar mass ($M_B$) of the acid:
\[ 0.17 = 1 \times 1.7 \times \frac{1.22 \times 1000}{M_B \times 100} \]
\[ 0.17 = 1.7 \times \frac{12.2}{M_B} \implies 0.1 = \frac{12.2}{M_B} \implies M_B = 122 \text{ g mol}^{-1} \]
2. Now, calculate $\Delta T_b$ for the benzene solution ($i = 0.5$ for complete dimerization):
\[ \Delta T_{b, \text{benz}} = 0.5 \times 2.6 \times \frac{1.22 \times 1000}{122 \times 100} \]
\[ \Delta T_{b, \text{benz}} = 0.5 \times 2.6 \times 0.1 = 0.13^\circ\text{C} \]
3. Comparing with $x \times 10^{-2}$:
\[ 0.13 = 13 \times 10^{-2} \implies x = 13 \] Step 4: Final Answer:
The value of $x$ is 13.
