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Question

The volume of 2.8 g of carbon monoxide at 27 C and 0.821 atm is

A
30 L
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B
3 L
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C
0.3 L
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D
1.5 L
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Solution

The correct option is B 3 L
According to the ideal gas equation, we have
PV=nRT
PV=(wM)RT
V=(wM)(RTP)
Given values are:
w = 2.8 g
M = Molar mass of CO = 28 g mol1
T = 27 C = (273+27) K = 300 K
P = 0.821 atm
R=0.0821 L atm mol1 K1
Putting the values in the formula we get :
V=(2.8 g28 g)mol1×(0.0821 L atm mol1K1)×(300 K)(0.821 atm)
=3L

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