The concentration of carbon monoxide in a sample of air is 8.1×10−6. there are molecules of co in 1.00 l of this air at 755 torr and 23 ∘c.

2.0*

Explanation:

Let us calculate total moles (air + CO)

moles (n) = PV /RT

P = 755 torr/760 = 0.9934 atm

V = 1.00 L

R = gas constant

T = 23°C + 273 = 296 K

So,

total moles (n) = 0.9934 x 1.00/0.0821 x 296 = 0.041 moles

mole fraction of CO = 8.1 x

moles of CO =8.1 x *0.041

                     =3.31 x moles

number of molecule of CO =3.31 x *(6.0233x )

                     = 2.0*

its a 5.5 n

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