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 \(0.586 \mathrm{~g}\) helium? (At \(25^{\circ} \mathrm{C}\) the vapor pressure of water is \(23.8\) torr.)

Short Answer

Expert verified
The total volume of gas collected to obtain \(0.586 \mathrm{~g}\) helium is approximately \(0.03632 \mathrm{~L}\).

Step by step solution

01

Write down the given information

We have the following information provided in the problem: - Mass of Helium (He): \(0.586 \mathrm{~g}\) - Temperature (T): \(25^{\circ} \mathrm{C}\) or \(298.15 \mathrm{~K}\) (converting Celsius to Kelvin by adding 273.15) - Total pressure (P_total): \(1.00 \mathrm{~atm}\) - Vapor pressure of water (P_water): \(23.8 \mathrm{~torr}\)
02

Convert pressure units to make them consistent

Both the total pressure and vapor pressure of water should be in the same unit. Since the vapor pressure of water is given in torr, let's convert the total pressure to torr. We know that 1 atm is equal to 760 torr. Therefore, the conversion can be done as follows: $$ P_{total} = 1.00 \mathrm{~atm} \times \frac{760 \mathrm{~torr}}{1 \mathrm{~atm}} = 760 \mathrm{~torr}$$
03

Calculate the partial pressure of Helium using Dalton's Law of Partial Pressures

According to Dalton's Law, the total pressure of a gas mixture is the sum of the partial pressures of the component gases. So, we can calculate the partial pressure of helium (P_He) as follows: $$ P_{He} = P_{total} - P_{water} $$ Substitute the given values: $$ P_{He} = 760 \mathrm{~torr} - 23.8 \mathrm{~torr} = 736.2 \mathrm{~torr} $$
04

Convert the mass of Helium to moles using its molar mass

The molar mass of helium is approximately 4.00 g/mol. We can calculate the moles of helium (n_He) as follows: $$ n_{He} = \frac{\text{Mass of Helium}}{\text{Molar Mass of Helium}} $$ Substitute the given value and molar mass: $$ n_{He} = \frac{0.586 \mathrm{~g}}{4.00 \mathrm{~g/mol}} = 0.1465 \mathrm{~mol} $$
05

Use the Ideal Gas Law to find the volume

The Ideal Gas Law equation is given by: $$ PV = nRT $$ We want to find the volume (V) of the helium gas, so we can rearrange the equation as follows: $$ V = \frac{nRT}{P} $$ Let's substitute the values of moles of helium (n_He), partial pressure of helium (P_He), gas constant (R) in L.torr/(mol.K), and temperature (T): $$ V = \frac{0.1465 \mathrm{~mol} \times 62.364 \frac{\mathrm{L\cdot torr}}{\mathrm{mol\cdot K}} \times 298.15 \mathrm{~K}}{736.2 \mathrm{~torr}} $$ Now, we can calculate the volume (V) of the helium gas: $$ V \approx 0.03632 \mathrm{~L} $$ So, the total volume of gas collected to obtain \(0.586 \mathrm{~g}\) helium is approximately \(0.03632 \mathrm{~L}\).

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Most popular questions from this chapter

An important process for the production of acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}\right)\) is given by the following equation: \(2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) A \(150 .-\mathrm{L}\) reactor is charged to the following partial pressures at \(25^{\circ} \mathrm{C}\) : $$ \begin{aligned} P_{\mathrm{C}_{3} \mathrm{H}_{6}} &=0.500 \mathrm{MPa} \\ P_{\mathrm{NH}_{3}} &=0.800 \mathrm{MPa} \\ P_{\mathrm{O}_{2}} &=1.500 \mathrm{MPa} \end{aligned} $$ What mass of acrylonitrile can be produced from this mixture \(\left(\mathrm{MPa}=10^{6} \mathrm{~Pa}\right) ?\)

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