Answer:
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Answer:
uh A.)? [ZnCI2+H2]? /////////
1. The molar mass of the unknown gas obtained is 0.096 g/mol
2. The pressure of the oxygen gas in the tank is 1.524 atm
<h3>Graham's law of diffusion </h3>
This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e
R ∝ 1/ √M
R₁/R₂ = √(M₂/M₁)
<h3>1. How to determine the molar mass of the gas </h3>
- Rate of unknown gas (R₁) = 11.1 mins
- Rate of H₂ (R₂) = 2.42 mins
- Molar mass of H₂ (M₂) = 2.02 g/mol
- Molar mass of unknown gas (M₁) =?
R₁/R₂ = √(M₂/M₁)
11.1 / 2.42 = √(2.02 / M₁)
Square both side
(11.1 / 2.42)² = 2.02 / M₁
Cross multiply
(11.1 / 2.42)² × M₁ = 2.02
Divide both side by (11.1 / 2.42)²
M₁ = 2.02 / (11.1 / 2.42)²
M₁ = 0.096 g/mol
<h3>2. How to determine the pressure of O₂</h3>
From the question given above, the following data were obtained:
- Volume (V) = 438 L
- Mass of O₂ = 0.885 kg = 885 g
- Molar mass of O₂ = 32 g/mol
- Mole of of O₂ (n) = 885 / 32 = 27.65625 moles
- Temperature (T) = 21 °C = 21 + 273 = 294 K
- Gas constant (R) = 0.0821 atm.L/Kmol
The pressure of the gas can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
Divide both side by V
P = nRT / V
P = (27.65625 × 0.0821 × 294) / 438
P = 1.524 atm
Learn more about Graham's law of diffusion:
brainly.com/question/14004529
Learn more about ideal gas equation:
brainly.com/question/4147359
By use of combined gas law
P2= T2P1V1/V2
v1=5L
P1=540 torr
T1=25+273=298k
V2=15 L
T2=32+273=305 k
P2 is therefore=( 305k x 540 torr x5 L) /( 15L x 298)= 184.23 torr
