How many grams of nitrogen gas are present in a 65.0 L gas cylinder at 25oC and 1 atm pressure?
PV = nRT
1×65.0 = n× 8.314 J/mol·K×298
65= n×2477.572
n= 65/2477.572
n= 0.026
Molar mass of nitrogen = 14.0067
= 14.0067×0.026=0.3675g
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