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Dr. I. M. A. Brightguy adds 0.1727 g of an unknown gas to a 125-mL flask. If Dr. B finds the pressure to be 736 torr at 20.0°C, is the gas likely to be methane, CH4, nitrogen, N2, oxygen, O2, neon, Ne, or argon, Ar?

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anfabba15

Answer:

The gas that Dr. Brightguy added was O₂

Explanation:

Ideal Gases Law to solve this:

P . V = n . R . T

Firstly, let's convert 736 Torr in atm

736 Torr is atmospheric pressure = 1 atm

20°C = 273 + 20 = 293 T°K

125 mL = 0.125L

0.125 L . 1 atm = n . 0.082 L.atm / mol.K . 293K

(0.125L .1atm) / (0.082 mol.K /L.atm . 293K) = n

5.20x10⁻³ mol = n

mass / mol = molar mass

0.1727 g / 5.20x10⁻³ mol = 33.2 g/m

This molar mass corresponds nearly to O₂

Lanuel

A gas with a molar mass of 33.21 g/mol is likely to be oxygen ([tex]O_2[/tex]) because the molar mass of oxygen ([tex]O_2[/tex]) is 32.00 g/mol.

Given the following data:

  • Mass = 0.1727 g
  • Volume = 125-mL.
  • Temperature = 20.0°C
  • Pressure = 736 torr
  • Ideal gas constant, R = 0.0821L⋅atm/mol⋅K
  • Ideal gas constant, R = 0.0821L⋅atm/mol⋅K

Conversion:

  • Volume = 125-mL to Liters = 0.125 L.
  • Temperature = 20.0°C to Kelvin = 293 K.
  • Pressure = 736 torr to atm = 1 atm.

To find how many moles (number of moles) of unknown gas are in a 125-mL flask, we would use the ideal gas law equation;

[tex]PV = nRT[/tex]

Where;

  • P is the pressure.
  • V is the volume.
  • n is the number of moles of substance.
  • R is the ideal gas constant.
  • T is the temperature.

Making n the subject of formula, we have;

[tex]n = \frac{PV}{RT}[/tex]

Substituting the given parameters into the formula, we have;

[tex]n = \frac{1(0.125)}{0.0821(293)} \\\\n = \frac{0.125}{24.06}[/tex]

Number of moles, n = 0.0052 moles.

Now, we can determine the unknown gas:

[tex]Molar \;mass = \frac{Mass}{number\;of\;moles}\\\\Molar \;mass = \frac{0.1727}{0.0052}[/tex]

Molar mass = 33.21 g/mol.

Therefore, a gas with a molar mass of 33.21 g/mol is likely to be oxygen ([tex]O_2[/tex]) because the molar mass of oxygen ([tex]O_2[/tex]) is 32.00 g/mol.

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