Gustina izvjesnog gasovitog ugljovodonika na 25 C iznosi 12.20 g/dm-3 i pritisku 1013k Pa i 5.900 g/dm-3 na pritisku 506.5 kPa.Izracunati molarnu masu gasa i dati njegovu vjerovatnu formulu.
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Pomocu opste jednacine
Pomocu opste jednacine gasnog stanja odredi molarnu masu jedinjenja:
pV=nRT
pV=(m/M)RT
pM=(m/V)RT
pM=ro*RT----->M=ro*R*T/p
Pazi na jedinice (ro=g/m3, moras da pretvoris, i T u Kelvine i p u Pa).
M1=30 g/mol; M2=29 g/mol
Msr=(M1+M2)/2= 29.5 g/mol, sto je priblizno 30 g/mol
Formula: CxHy, x moze biti 2 (2*12=24; 30-24=6), sto nam ostavlja 6 atoma vodonika. Dakle, formula je C2H6, a gas je etan.
Pozdrav! :)
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