How many molecules of N2 are contained in 1.12 L of air if the volume content of N2 in air is 80% at NTP.
We can use the ideal gas law in solving the amount of molecules present in a volume sample of air.
`PV = nRT`
We know that the volume sample is not pure N_2 therefore we can label it as n_total.
`PV = n_(t otal) RT`
Then arranging the expression we can have:
`n_(t otal) = (PV)/(RT)`
At NTP (or Normal Temperature and Pressure), the values of T and P are 203.15 K and 1.00 atm respectively. The value of gas constant (R) to be used is `0.08206 (atm-L)/(mol-K)` . The volume is measured to be 1.12 L.
`n_(t otal) =(PV)/(RT)`
`n_(t otal)=(1*1.12L)/(0.08206 *293.15)`
`n_(t otal)= 0.046558 mol es of air sampl e`
We know that 80% of the total sample is `N_2` so we can get the amount of moles of `N_2` by:
`0.046558 t otal mol es * (80/100)`
`= 0.037247 mol es N_2`
Finally, to get the number of molecules, use the Avogadro's number.
`0.037247 mol es N_2 * (6.022 x10^23 mol ec u l e s)/(1 mol e N_2)`
`= 2.243x10^22 mol ec u l es of N_2` -> final answer
Now the percentage of N2 in air is 78 % that is there are 78 mL of N2 in 100m L of air
so the total volume of N2 in 1.12 L would be
1.12 L * (1000 mL air/ 1L ) *(78 mL N2/100 mL of air)
873.6 mL of N2 that is 0.87 L of N2
Now at STP the volume fo 1 mole of any gas is 22.4 L and we know that
1 mole = (6.023*10^23 molecule)
so 0.87 L of N2 would have
0.87 L N2 * (1 mol N2/22.4 L N2 )*(6.023*10^23 molecule of N2/ 1mol N2)
2.34*10^22 molecule of N2 is present.