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Fiesta28 [93]
2 years ago
9

the density of oxygen 1.43 gm/liter at 0°c and pressure 1.0 atm. if a 20 liter cylinder is filled with oxygen at pressure of 25

atm and temperature of 27°c. what is the mass of oxygen in the cylinder
Chemistry
1 answer:
lbvjy [14]2 years ago
6 0

Answer:

640 g

Explanation:

Step 1: Given and required data

  • Volume of the cylinder (V): 20 L
  • Pressure of the oxygen (P): 25 atm
  • Temperature (T): 27 °C (300 K)
  • Ideal gas constant (R): 0.082 atm.L/mol.K

Step 2: Calculate the moles of oxygen gas

We will use the ideal gas equation

P × V = n × R × T

n = P × V / R × T

n = 25 atm × 20 L / (0.082 atm.L/mol.K) × 300 K = 20 mol

Step 3: Calculate the mass corresponding to 20 moles of oxygen

The molar mass of oxygen is 32.00 g/mol.

20 mol × 32.00 g/mol = 640 g

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Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

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Mole fraction of 2‑Propanol (x_2) = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

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p_1 = partial vapor pressure of 1‑Propanol

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p_2=x_2\times p^o_2

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p_2 = partial vapor pressure of 2‑Propanol

p^o_2 = vapor pressure of pure substance 2‑Propanol

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Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352

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\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

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