A mixture of helium and neon gases is collected over water at $28.0 \rm~^\circ C$ and $745\rm~ mmHg$. If the partial pressure of helium is $368\rm~ mmHg$, what is the partial pressure of neon? (Vapor pressure of water at $28\rm~^\circ C =28.3~mmHg$). Calculate the weight percentage of He and Ne in the original mixture.
The question is based on Dalton's law of partial pressure. I got the first part right.In the second part, since we have to deal only with the original mixture, total pressure should be $$745-28.7=716.7\rm~mm Hg$$ So we calculate the total moles in terms of "v" (total volume) and then moles of neon and helium in terms of "v" from which we calculate the weight of neon and helium and calculate the weight percentage. "v" will eventually cancel out and we get weight percentage of neon as 34.54 % and that of helium as 65.45 % but that is not the right answer.In the solution to this question, they took the total pressure while dealing with the second part as 748 mm hg only..but why would we do that.We are only talking about the original mixture. Where am I wrong?
Answer in my textbook : Neon weight % = 82.58 %, helium weight% : 17.43%