The total vapour pressure of a 4 mole % solution of `NH_(3)` in water at 293k is 50.0 torr, the vapour pressure of pure water is 17.0 torr at this temperature. Applying Henry's and Raoult's laws, calculate the total vapour pressure for a 5 mole% solution

Options

A. 58.25torr
B. 33torr
C. 42.1 torr
D. 52.25 torr

Answer

A

Explanation

Given: `p_("water")^(@) = 17.0` torr, `P_("total")` (4mole % solution) `= P_(NH_(3)) + P_("water") = 50.0` torr
`x_(NH_(3)) = 0.04 and x_("water") = 0.96`
According to Raoult.s law, `P_("water") = x_("water") P_("water")^(@) = 0.96 xx 17.0` torr= 16.32 torr
Henry.s law constant for ammonia is `K_(H) (NH_(3))=(P_(NH_(3)))/(x_(NH_(3)))=(33.68 "torr")/(0.04)= 842` torr
Hence, for 5 mole % solution, we have
`P_(NH_(3))= K_(H) (NH_(3)) x_(NH_(3)) = (842 "torr") (0.05)= 42.1` torr
`P_("water") = P_("water")^(@) x_("water")` =(17 torr) (0.95)= 16.15 torr
Thus, `P_("total")` = (5 mole % solution) =`P_(NH_(3)) + P_("water")= 42.1 + 16.15 = 58.25` torr