n=(755-23.8)*.500/(8.31*298)= 0.147634854 moles nitrogen gas
Now, at STP, now water vapor, that translates tso
V= nRT/P= 0.147634854*8.13*273/(755-23.8)=.448 dm^3 so I agree with you.
Exactly 500 cm3 of nitrogen is collected over water at 25°C and 755 mmHg. The gas is saturated with water
vapour. Calculate the volume of the nitrogen in the dry condition at 0 °C and 1 atm. Vapour pressure of water at
25°C is 23.8 mmHg.
My working is:
P =1.0247 atm (converting mmHg to atm), V=0.5L, T=323K (25+25+273), R=0.08206.
Using PV=nRT, I solved for n, which is 0.01933. Using this formula again, with the other info, 1atm(V)=0.01933(0.08206)(273K) and I get V=433 cm^3. The answer says 441cm^3 and I don't know what I could have possibly done wrong. Any help is much appreciated.
1 answer