If 250 mL of `N_(2)` over water at `30^(@)C` and a total pressure of 740 torr is mixed with 300 mL of Ne over water at `25^(@)C` and a total pressure of 780 torr, what will be the total pressure if the mixture is in a 500 mL vessel over water at `35^(@)C`. (Given : Vapour pressure (Aqueous tension )of `H_(2)O` at `25^(@)C` and `35^(@)C` are 23.8, 31.8 and 42.2 torr respectively. Assume volume of `H_(2)O(l)` is negligible in final vessel)A. 760 torrB. 828.4 torrC. 807.6 torrD. 870.6 torr

If 250 mL of `N_(2)` over water at `30^(@)C` and a total pressure of 7

1.	If 250 mL of `N_(2)` over water at `30^(@)C` and a total pressure of 740 torr is mixed with 300 mL of Ne over water at `25^(@)C` and a total pressure of 780 torr, what will be the total pressure if the mixture is in a 500 mL vessel over water at `35^(@)C`. (Given : Vapour pressure (Aqueous tension )of `H_(2)O` at `25^(@)C` and `35^(@)C` are 23.8, 31.8 and 42.2 torr respectively. Assume volume of `H_(2)O(l)` is negligible in final vessel)A. 760 torrB. 828.4 torrC. 807.6 torrD. 870.6 torr
Answer» Correct Answer - d (d) `n_(N_(2))=(((708.2)/(760)xx0.25))/(0.0821xx303)=9.36xx10^(-3)` `n_(o_(2))=(((756.2)/(760))xx0.3)/((0.0821)xx298)` `=0.0122` `n_("total")"moles"=0.02156` Pressure in final vessel = P `((n_("total"))RT)/(V)=(0.02156xx00821xx308)/(0.5)` P=1.09 atm or 828.4 torr `P_("total")=P_((O_(2)+N_(2)))+V.pr. "of "H_(2)O` `=828.4+42.2=870.6"torr" `

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