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Chapter 5: Problem 100

Helium is collected over water at \(25^{\circ} \mathrm{C}\) and 1.00 atm total pressure. What total volume of gas must be collected to obtain 0.586 g helium? (At \(25^{\circ} \mathrm{C}\) the vapor pressure of water is 23.8 torr.)

Short Answer

Expert verified
To obtain 0.586 g of helium at \(25^{\circ} \mathrm{C}\) and 1.00 atm total pressure, the total volume of helium gas collected over water should be approximately 3.90 L.

Step by step solution

01

Convert the given information to appropriate units

Before we begin, let's first convert the given information to the appropriate units for the Ideal Gas Law equation, which is PV = nRT. Temperature = \(25^{\circ} \mathrm{C}\) = (25 + 273.15) K = 298.15 K Total pressure = 1.00 atm (Already in the correct unit) Vapor pressure of water = 23.8 torr. We need to convert this to atm. \(1 \mathrm{atm} = 760 \mathrm{torr}\) Water vapor pressure = \(\frac{23.8 \mathrm{torr}}{760 \mathrm{torr/atm}} = 0.03132 \mathrm{atm}\) Helium mass = 0.586 g Molar mass of helium = 4.00 g/mol
02

Calculate the partial pressure of helium

Since helium is collected over water, we need to find the partial pressure of helium. The total pressure is the sum of the pressure due to helium and the pressure due to water vapor. Total pressure = Helium pressure + Water vapor pressure Helium pressure = Total pressure - Water vapor pressure \(P_{\mathrm{He}} = 1.00 \mathrm{atm} - 0.03132 \mathrm{atm}\) \(P_{\mathrm{He}} = 0.96868 \mathrm{atm}\)
03

Calculate the moles of helium collected

Next, we need to compute the number of moles of helium collected. To do this, we will use its mass and molar mass: Moles of helium = \(\frac{\mathrm{mass}}{\mathrm{molar \, mass}}\) n = \(\frac{0.586 \, \mathrm{g}}{4.00 \mathrm{g/mol}}\) n ≈ 0.1465 mol
04

Use the Ideal Gas Law to find the volume

Now that we have the partial pressure of helium, the temperature, and the moles of helium, we can use the Ideal Gas Law to find the volume of gas collected. \(PV = nRT\) \(V = \frac{nRT}{P}\) V = \(\frac{(0.1465 \, \mathrm{mol})(0.0821 \, \mathrm{L \cdot atm/mol \cdot K})(298.15 \, \mathrm{K})}{0.96868 \, \mathrm{atm}}\) V ≈ 3.90 L
05

Write the final answer

To obtain 0.586 g of helium at \(25^{\circ} \mathrm{C}\) and 1.00 atm total pressure, the total volume of helium gas collected over water should be approximately 3.90 L.

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Most popular questions from this chapter

A certain flexible weather balloon contains helium gas at a volume of 855 L. Initially, the balloon is at sea level where the temperature is $25^{\circ} \mathrm{C}$ and the barometric pressure is 730 torr. The balloon then rises to an altitude of 6000 ft, where the pressure is 605 torr and the temperature is \(15^{\circ} \mathrm{C}\). What is the change in volume of the balloon as it ascends from sea level to 6000 ft?

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