Question

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.

Answer 1

2.0*
`10^(17)`

Step-by-step 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
`10^(-6)`

moles of CO =8.1 x
`10^(-6)` *0.041

=3.31 x
`10^(-7)` moles

number of molecule of CO =3.31 x
`10^(-7)` *(6.0233x
`10^(23)`)

= 2.0*
`10^(17)`