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Chapter 12: Problem 35

A 25.0 L volume of \(\mathrm{He}(\mathrm{g})\) at \(30.0^{\circ} \mathrm{C}\) is passed through \(6.220 \mathrm{g}\) of liquid aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\) at \(30.0^{\circ} \mathrm{C} .\) The liquid remaining after the experiment weighs \(6.108 \mathrm{g}\) Assume that the He(g) becomes saturated with aniline vapor and that the total gas volume and temperature remain constant. What is the vapor pressure of aniline at \(30.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The vapor pressure of aniline at \(30.0^{\circ} C\) is \(0.012 atm\)

Step by step solution

01

Determine the amount of aniline that has vaporized

From the given weights, we can find the amount of aniline that has vaporized by subtracting the final weight of the remaining liquid aniline from the initial weight, which yields \(6.220g - 6.108g = 0.112g\) aniline vapor.
02

Convert the weight of aniline vapor to moles

Next, we will convert this weight of aniline vapor into moles, using the molar mass of aniline which is about \(93.13g/mol.\) Hence, the number of moles of aniline vapor is \(0.112g ÷ 93.13g/mol = 1.202 x 10^{-3} moles\).
03

Apply the Ideal Gas law

The ideal gas law states that P = nRT/V. Here, P is the vapor pressure (which we are looking for), n is the number of moles of aniline in the gaseous state, R is the gas constant (0.0821 Latm/mol.K for this problem), T is the temperature in Kelvin, and V is the volume. However, it is crucial to convert the given temperature from Celsius to Kelvin by adding 273.15 to the Celsius temperature. Thus, T = 30°C + 273.15 = 303.15 K.
04

Calculate the vapor pressure of aniline

Substitute the known values into the ideal gas law to solve for the vapor pressure P. P = (1.202 x 10^{-3} mol * 0.0821 L.atm/mol.K * 303.15K) / 25.0L. Solving for P gives \(P = 0.012 atm\)

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Most popular questions from this chapter

Sketch a plausible phase diagram for hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) from the following data: triple point \(\left(2.0^{\circ} \mathrm{C}\right)\) and \(3.4 \mathrm{mm} \mathrm{Hg}\) ), the normal melting point \(\left(2^{\circ} \mathrm{C}\right),\) the normal boiling point \(\left(113.5^{\circ} \mathrm{C}\right),\) and the critical point \(\left(380^{\circ} \mathrm{C} \text { and } 145 \mathrm{atm}\right) .\) The density of the liquid is less than that of the solid. Label significant data points on this diagram. Are there any features of the diagram that remain uncertain? Explain.

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