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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)

Text Solution

Verified by Experts

The correct Answer is:
870.6
`n_(N_(2)) = ((708.2/760 xx 0.25)/(0.0821 xx 303)) = 9.36 xx 10^(-3)`
`n_(N_(2)) = ((756.2/760) xx 0.25)/(0.0821 xx 298)`
=0.0122
`n_("total") "moles" = 0.02156`
`(n_("total")RT)/V = (0.02156 xx 0.0821 xx 308)/0.5`
p = 1.09 atm or 828.4 torr
`P_("total") = p_(N_(2) + N_(e)) + V.p. "of" H_(2)O`
`=828.4 + 42.4 = 870.6` torr
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