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Glycerin Solution Vapor Pressure at 39.88°C
Chapter 11, Problem 51(choose chapter or problem)
Glycerin, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\), is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 164 g glycerin to 338 mL \(\mathrm{H}_{2} \mathrm{O}\) at \(39.8^{\circ} \mathrm{C}\)? The vapor pressure of pure water at \(39.8^{\circ} \mathrm{C}\) is 54.74 torr and its density is \(0.992 \mathrm{g} / \mathrm{cm}^{3}\) .
Questions & Answers
QUESTION:
Glycerin, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\), is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 164 g glycerin to 338 mL \(\mathrm{H}_{2} \mathrm{O}\) at \(39.8^{\circ} \mathrm{C}\)? The vapor pressure of pure water at \(39.8^{\circ} \mathrm{C}\) is 54.74 torr and its density is \(0.992 \mathrm{g} / \mathrm{cm}^{3}\) .
ANSWER:Step 1 of 2
Given:
\(\begin{array}{l}\text{Mass of glycerin } \left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right)=164 \mathrm{~g}\\ \text{Volume of water } \left(\mathrm{H}_{2} \mathrm{O}\right)=338 \mathrm{~mL}\\ \text{Vapor pressure of pure water } \left(\mathrm{P}_{0}\right) \text{ at } 39.88^{\circ} \mathrm{C}=54.74 \text{ torr}\\ \text{Density of water at } 39.88^{\circ} \mathrm{C}=0.992 \mathrm{~g} / \mathrm{cm}^{3}\end{array}\)
Let's first calculate the mass of water using the given volume and density:
\text { Mass of water }=\text { Density of water } \times \text { Volume of water }=0.992 \mathrm{g} / \mathrm{cm}^{3} \times 338 \mathrm{cm}^{3}=335.3 \mathrm{g}\)
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Glycerin Solution Vapor Pressure at 39.88°C
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Discover how to calculate the vapor pressure of a glycerin solution when mixed with water at a specific temperature. A valuable chemistry lesson.