Chapter 2 Solutions Class 12 Chemistry Maharashtra Board Textbook Solution
11. A solution of citric acid C6H8O7 in 50 g of acetic acid has a boiling point elevation
of 1.76 K. If Kb for acetic acid is 3.07 K kg mol-1, what is the molality of solution?
(0.573 m)
Answer:
Given: Boiling point elevation (\(\Delta T_b\)) = 1.76 K
Kb of acetic acid = 3.07 K kg mol-1
Mass of acetic acid = 50 g
To find: Molality of the solution
Formula: \(\Delta T_b = K_b \cdot m\)
Calculation: Using the formula and rearranging, we get,
\[ m = \frac{\Delta T_b}{K_b} = \frac{1.76 \, \text{K}}{3.07 \, \text{K kg}^{-1}} = 0.573 \, \text{mol kg}^{-1} \]
The molality of the solution is 0.573 m.
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