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Step-by-Step Solution
Step 1: Identify the Given Data
We have two solvents, A and B, into which the same amount of a non-volatile, non-electrolyte solute (1 g) is dissolved in each case. The mass of solvent is 100 g for each case, and the ratio of their ebullioscopic constants (molal elevation constant), $k_b(A):k_b(B)$, is given as 1 : 5.
Step 2: Recall the Formula for Elevation in Boiling Point
The elevation in boiling point, $\Delta T_b$, for a solution is given by:
$$ \Delta T_b = k_b \times \text{molality} $$
Here, $k_b$ is the ebullioscopic constant of the solvent, and the molality is defined as moles of solute per kilogram of solvent.
Step 3: Note that the Molality is the Same for Both Solutions
Since the same mass (1 g) of the same solute is dissolved in the same mass of solvent (100 g) in both cases, the molality for solvent A and solvent B will be the same (denoted by $m$) because the amount of solute and mass of solvent are identical in each case.
Step 4: Calculate the Ratio of the Elevations
The ratio of elevations in boiling point for the two solvents can be written as:
$$ \frac{\Delta T_b(A)}{\Delta T_b(B)} \;=\; \frac{k_b(A) \times m}{k_b(B) \times m} \;=\; \frac{k_b(A)}{k_b(B)} $$
We are given:
$$ \frac{k_b(A)}{k_b(B)} = \frac{1}{5} $$
Step 5: State the Final Result
Hence, the ratio of the elevation of boiling points is:
$$ \frac{\Delta T_b(A)}{\Delta T_b(B)} \;=\; \frac{1}{5} $$
Answer
The ratio of their elevation in boiling points is 1 : 5.
