QQuestionAnatomy and Physiology
QuestionAnatomy and Physiology
What is the freezing point of vinegar, which is a $5.00 \%$ (by mass) solution of acetic acid $\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)$ in water? The dissociation constant $\left(K_{a}\right)$ for acetic acid is $1.80 \times 10^{- 5}$; the density of the solution is $1.006 \mathrm{~g} / \mathrm{mL}$. For water, the cryoscopic constant $\left(K_{f}\right)$ is $1.86 \backslash, \wedge \backslash$ circ $\backslash$ text $[\mathrm{C} / \mathrm{m}]$.
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Step 1Let'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
Final Answer
The freezing point of the vinegar solution is - 1.63 \mathrm{~°C}.
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