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Step-by-Step Solution
Step 1: Understanding the problem
We have 1.22 g of an organic acid dissolved in two different solvents: benzene and acetone. The acid dimerizes (forms pairs) in benzene and stays in monomeric form in acetone. We are given:
K_b (benzene) = 2.6 K·kg·mol−1
K_b (acetone) = 1.7 K·kg·mol−1
The boiling point elevation in acetone ( \Delta T_b ) = 0.17 °C
Solvent mass in each case = 100 g = 0.1 kg
We need to find the elevation in boiling point in benzene, expressed as x \times 10^{-2} °C, and then determine the value of x .
Step 2: Recall the relevant formula
Elevation in boiling point ( \Delta T_b ) for a solution is given by:
\Delta T_b = i \; K_b \; m
where:
i is the Van’t Hoff factor (the effective number of particles per formula unit in solution).
K_b is the ebullioscopic constant (or boiling point elevation constant) of the solvent.
m is the molality of the solution (moles of solute per kilogram of solvent).
Step 3: Determine the Van’t Hoff factors in each solvent
1. In acetone, the acid remains a monomer. Thus, there is no association or dissociation, which means:
i (\text{acetone}) = 1.
2. In benzene, the acid dimerizes fully. Two acid molecules combine to form one dimer, effectively halving the number of particles. Therefore,
i (\text{benzene}) = \frac{1}{2}.
Step 4: Determine the molality using acetone data
From the elevation in boiling point in acetone:
\Delta T_b (\text{acetone}) = 0.17 = i \times K_b(\text{acetone}) \times m.
Substitute i = 1 and K_b(\text{acetone}) = 1.7 :
0.17 = \; 1 \times 1.7 \times m.
So,
m = \frac{0.17}{1.7} = 0.1 \text{ mol kg}^{-1}.
But by definition,
m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}.
Since the solvent mass is 0.1 kg (100 g), the moles of solute in 1.22 g must be:
\text{moles of solute} = m \times \text{kg of solvent} = 0.1 \times 0.1 = 0.01.
Hence,
\text{moles of solute} = 0.01.
From moles = (mass in grams)/(molar mass), we get:
0.01 = \frac{1.22}{M_w} \quad \Rightarrow \quad M_w = \frac{1.22}{0.01} = 122 \text{ g/mol}.
Step 5: Calculate the boiling point elevation in benzene
Now we use the same approach for benzene. Since the acid dimerizes, i = \frac{1}{2} . Thus:
\Delta T_b (\text{benzene}) = \left(\frac{1}{2}\right) \times K_b(\text{benzene}) \times m_\text{benzene}.
We have:
K_b(\text{benzene}) = 2.6
The molality m_\text{benzene} is given by
m_\text{benzene} = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{1.22 / 122}{0.1} = 0.1 (same total moles, same mass of solvent).
Therefore:
\Delta T_b (\text{benzene}) = \frac{1}{2} \times 2.6 \times 0.1 = \frac{2.6}{2} \times 0.1 = 1.3 \times 0.1 = 0.13 \text{ °C}.
Step 6: Express the answer in the required form
We need to express the elevation in boiling point for benzene as x \times 10^{-2} °C. Since
\Delta T_b (\text{benzene}) = 0.13 \text{ °C} = 13 \times 10^{-2} \text{ °C},
we have
x = 13.
Final Answer
The value of x is 13.
