The mass of N₂ in the mixture having a total pressure of 784 torr with an H₂ partial pressure of 124 torr is 1.988 g
We'll begin by calculating the partial pressure of N₂. This can be obtained as follow:
Total pressure = 784 torr
Partial pressure of H₂ = 124 torr
<h3>Partial pressure of N₂ =?</h3>
Total pressure = Partial pressure of H₂ + Partial pressure of N₂
784 = 124 + Partial pressure of N₂
Collect like terms
Partial pressure of N₂ = 784 – 124
<h3>Partial pressure of N₂ = 660 Torr</h3>
- Next, we shall determine the number of mole of N₂ in the mixture.
Pressure of N₂ (P) = 660 Torr
Volume (V) = 2 L
Temperature (T) = 298 K
Gas constant (R) = 62.364 L.Torr/Kmol
<h3>Number of mole of N₂ (n) =? </h3>
PV = nRT
660 × 2 = n × 62.364 × 298
1320 = n × 18584.472
Divide both side by 18584.472
n = 1320 / 18584.472
<h3>n = 0.071 mole</h3>
- Finally, we shall determine the mass of N₂ in the mixture.
Mole of N₂ = 0.071 mole
Molar mass of N₂ = 2 × 14 = 28 g/mol
<h3>Mass of N₂ =? </h3>
Mass = mole × molar mass
Mass of N₂ = 0.071 × 28
<h3>Mass of N₂ = 1.988 g</h3>
Therefore, the mass of N₂ in the mixture having a total pressure of 784 torr with an H₂ partial pressure of 124 torr is 1.988 g
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