Calculate the number of molecules present in 350 `cm^(3)` of `NH_(3)` gas at 273 K ans 2 atmosphere pressure.

Explanation

First of all, we have to determine the volume of the gas at STP.
`{:("Given conditions","At STP"),(V_(1)=350cm^(3),V_(2)=?),(T_(1)=273K,T_(2)=273K),(P_(1)="2 atmospheres",P_(2)="1 atm"):}`
`"Applying gas equation :"=(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)), " we get"(350xx2)/(273)=(1xxV_(2))/(272) or V_(2)=(350xx2)/(273)xx(273)/(1)=700cm^(3)`
By mole concept, 1 mole of `NH_(3)=6.022xx10^(23)" molecules"=22400 cm^(3)" at STP"`
Thus, 22400 `cm^(3)` of `NH_(3)` at STP contain `=(6.022xx10^(23))/(22400)xx700=1.882xx10^(22)" molecules"`
`"Alternatively, applying gas equation, PV=nRT"`
`n=(PV)/(RT)=("2 atm"xx0.350L)/("0.0821 L atm K"^(-1)"mol"^(-1)xx273K)=0.0312" mole"`
`therefore" No. of molecules"=0.0312 xx6.022xx10^(23)=1.88xx10^(22)`.