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Find the mass of 5.60 L O2 at 1.75 atm and 250.0 K

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Answer:

Moles O2 = pV/RT = 2.75 x 5.60 / 0.08206 x 250=0.751

mass = 32 g/mol x 0.751 mol= 24.0 g

Explanation:

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The mass of 5.60 Liter oxygen at 1.75 atm pressure and 250.0 K temperature is 24.0 gram.

What is ideal gas equation?

Ideal gas equation is a hypothesis which tells about the behavior of gas at standard condition and it is represented as:

PV = nRT, where

P = pressure = 1.75 atm

V = volume = 5.60 L

R = universal gas constant = 0.0821 L.atm /K.mol

T = temperature = 250 K

n is the moles of gas which will further define as:

n = W/M, where

W = required mass

M = molar mass

Now putting all the value in the above ideal gas equation and calculate for moles as:

n = (1.75)(5.60) / (0.0821)(250)

n = 0.751

Molar mass of oxygen is 32 g/mol and we calculate their required mass as:
W = 32 g/mol x 0.751 mol= 24.0 g

Hence, the required mass of oxygen is 24.0 g.

To know more about ideal gas equation, visit the below link:
https://brainly.com/question/25290815

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