How many molecules are there in an 11.2*L11.2L volume of gas at "NTP"NTP?

1 Answer
Jul 17, 2017

Approx. 3xx10^23*"molecules"3×1023molecules

Explanation:

As far as I know, and you will have to check your syllabus (do so, because this is important!), "NTP"NTP specifies a temperature of 293.15*K293.15K, and a pressure of 1*atm1atm. Unfortunately there seems to be several standards in use, "STP"STP, "NTP"NTP, and "SATP"SATP, and several "standard"standard pressures of 1*atm1atm, and 100*kPa100kPa; and this is bound to cause confusion, and error, and of course it does so. The given standards SHOULD be listed as supplementary material on each exam paper. You still have to be able to use the standards effectively.

So we use the Ideal Gas equation.......n=(PV)/(RT)n=PVRT

=(1*atmxx11.2*L)/(0.0821*(L*atm)/(K*mol)xx293.15*K)=0.465*mol=1atm×11.2L0.0821LatmKmol×293.15K=0.465mol.

We are still not done, because we were axed for THE NUMBER of molecules. And the molar quantity is by definition a NUMBER of particles.

1*"mole"-=6.022xx10^231mole6.022×1023 "individual particles"individual particles

And so......"Number of atoms"-=6.022xx10^23*mol^-1xx0.465*molNumber of atoms6.022×1023mol1×0.465mol

== "How many atoms?"How many atoms?

Capisce?