Question

1) A) A samole of flourine gas is in a 12.56 L cylinder at 41.35 degree celsius and 700 torres. What is the kinetic energy and the root mean square velocity of the flourine gas.

B) Methane gas diffuses at a rate of 32.25 ml/min. Under identical conditions an unknown gas diffuses at a rate of 22.28 ml/min. What is the molar mass of the unknown gas.

C) Oxygen IS produced by the decomposition of sodium bromate and collected over water. if 76.25 ml of wet oxygen is collected at 47.5 degree celsius with an atmospheric barometric pressure of 697.2 mmHg. How many milligrams of sodium bromate have been consumed?

D) Rank the following gases from most ideal to least ideal.

IF₃ CH₄ SF₆

Answer 1

1) Volume, V = 12.56 l

Temperature, T = 41.35 C = 41.35 +273.15 = 314.5 K

Pressure, P = 700 torr = 0.921 atm ( 1 torr = 0.001315 atm)

Number of moles, n = PV/RT = 0.921 atm x 12.56 lt/0.0821 lt atm/mol K x 314.5 K = 11.57 moles/25.82 = 0.448 moles

Kinetic energy, KE = (3/2) nRT = 3/2 x 0.448 moles x 8.314 J/mol K X 314.5 K = 1757.11 Joules = 1.757 kJ

Molar mass of fluorine = 37.997 g/mol = 0.037997 kg/mol

root mean square velocity of the flourine gas rms = 3RT/M = 3 x 8.314 Kg m²/s² mol K X 314.5 K/0.037997 kg/mol =

rms = 7844.26 m²/s² /0.037997 = 206444.167 m²/s² = 454.36 m/s

b) Diffusion of the gas r = 1/d

In the above formula d is density of gas whose formula where molecular weight is twice of vapor density (M= 2d) and already mentioned that identical conditions so remaining will be constant

But M (molar mass ) varies based on the gas so velocity of gas molecule is inversely proportional to the square root of molar mass

Molar mass of methane = CH₄ = 16.04 g/mol

r1/r2 = M2/M1 = 32.25 ml/min/22.28 ml/min = M2/16.04 g/mol

32.25 ml/min/22.28 ml/min = M2/16.04 g/mol

1.4475 = M2/16.04 g/mol

1.4475² = M2/16.04 g/mol

M2 = 2.095 x 16.04 g/mol = 33.61 g/mol

C) 2 NaBrO₃ ------------> 2NaBr + 3O₂

Atmospheric barometric pressure, Patm = 697.2 mm Hg

Volume of wet oxygen = 76.25 ml = 0.07625 atm

Temperature, T = 47.5 +273.15 = 320.65 K

Vapor pressure of water at 47.5 C = 89.5 mm Hg (from the chart)

Pressure of dry oxygen = 697.2 mm Hg - 89.5 mm Hg = 607.7 mm Hg = (760 mmHg = 1 atm) = 607.7 /760 = 0.7996 atm

Ideal gas equation : PV = nRT

0.7996 atm x 0.07625 lt = n x 0.0821 lt atm /mol K X 320.65 K

0.0609695 = n x 26.325 /mol

n = 0.0609695 mol/26.325 = 0.00232 moles

Number of moles of oxygen = 0.00232 moles

Molar ratio of sodium bromate and oxygen = 2:3

2 moles of sodium bromate ---> 3 moles of Oxygen produced

?? moles of sodium bromate ----> 0.00232 moles of oxygen

0.00232 x 2 /3 = 0.00155 moles of sodium bromate

Molar mass of sodium bromate = 150.89 g/mol ( 22.98 + 79.9 + (3x 15.99))

Mass of sodium bromate = Molar mass of sodium bromate x No. of moles = 150.89 g/mol x 0.00155 mol = 0.234 gms

4) Polar molecules have strong intermolecular attractions than non polar molecules whereas non polar molecules have only London dispersion forces.

Usually ideal gas molecules shouldn't have more intermolecular attractions

IF₃ : IF₃ is polar molecule and bond dipoles do not cancel

CH₄ : Intermolecular forces, ionic in character and non polar in nature. they produce only London dispersion forces and act as ideal behavior at the temperature when it is further from from boiling point

SF₆ : It is non polar molecule and bond dipoles cancel each other. They experience London dispersion forces.

CH₄ > SF₆ > IF₃