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Step 1:Let's solve this problem step by step:
Step 2:: Calculate the molality of the acetic acid solution
\text{Molality} = \frac{(5.00 \mathrm{~g})}{(60.052 \mathrm{~g/mol})(0.095 \mathrm{~kg})} = 0.873 \mathrm{~m}
- First, determine the mass of acetic acid and water in 100 g of solution - Calculate molality of acetic acid
Step 3:: Calculate the degree of dissociation
K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}]}{[\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}]} = \frac{x^{2}}{(0.873 - x)}
- Set up the dissociation equilibrium equation
Step 4:: Solve for degree of dissociation
x = \sqrt{(1.80 \times 10^{-5})(0.873)} = 0.00420 \mathrm{~m}
1.80 \times 10^{- 5} = \frac{x^{2}}{0.873}
Step 5:: Calculate freezing point depression
\Delta T_{f} = (1.86 \mathrm{~°C/m})(0.877 \mathrm{~m}) = 1.63 \mathrm{~°C}
- Freezing point depression formula:
Step 6:: Calculate freezing point
- Pure water freezes at $$0 \mathrm{~°C}
\text{Freezing point of solution} = 0 \mathrm{~°C} - 1.63 \mathrm{~°C} = - 1.63 \mathrm{~°C}
Final Answer
The freezing point of the vinegar solution is - 1.63 \mathrm{~°C}.
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