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

Answer:

2.0*$10^{17}$

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*$10^{17}$