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If 250 mL of N(2) over water at 30^(@)C...

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 torr

B

828.4 torr

C

807.6 torr

D

870.6 torr

Text Solution

Verified by Experts

The correct Answer is:
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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