Answer:
41.17g
Explanation:
We are given the following parameters for Flourine gas(F2).
Volume = 5.00L
Pressure = 4.00× 10³mmHG
Temperature =23°c
The formula we would be applying is Ideal gas law
PV = nRT
Step 1
We find the number of moles of Flourine gas present.
T = 23°C
Converting to Kelvin
= °C + 273k
= 23°C + 273k
= 296k
V = Volume = 5.00L
R = 0.08206L.atm/mol.K
P = Pressure (in atm)
In the question, the pressure is given as 4.00 × 10³mmHg
Converting to atm(atmosphere)
1 mmHg = 0.00131579atm
4.00 × 10³ =
Cross Multiply
4.00 × 10³ × 0.00131579atm
= 5.263159 atm
The formula for number of moles =
n = PV/RT
n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K
n = 1.0834112811moles
Step 2
We calculate the mass of Flourine gas
The molar mass of Flourine gas =
F2 = 19 × 2
= 38 g/mol
Mass of Flourine gas = Molar mass of Flourine gas × No of moles
Mass = 38g/mol × 1.0834112811moles
41.169628682grams
Approximately = 41.17 grams.
The mass of F₂ in the 5 L cylinder at 4×10³ mmHg and 23 °C is 41.154 g
Volume (V) = 5 L
Pressure (P) = 4×10³ mmHg
Temperature (T) = 23 °C = 23 + 273 = 296 K
Gas constant (R) = 62.364 mmHg.L/Kmol
PV = nRT
4×10³ × 5 = n × 62.364 × 296
20000 = n × 18459.744
Divide both side by 18459.744
n = 20000 / 18459.744
Mole of F₂ = 1.083 mole
Molar mass of F₂ = 2 × 19 = 38 g/mol
Mass = mole × molar mass
Mass of F₂ = 1.083 × 48
Therefore, the mass of F₂ in the cylinder is 41.154 g
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