At 25 oC, oxygen gas collected over water at a total pressure of 758 torr had a

Question

At 25 oC, oxygen gas collected over water at a total pressure of 758 torr had a concentration of 0.0393 g/L. What would its concentration be if its partial pressure over water were 803 torr? (Vapor pressure of water at 25 oC is 23.8 torr)?

A) 0.0360 g/L

B) 0.0743 g/L

C) 0.0680 g/L

D) 0.0500 g/L

E) 0.0430 g/L

A solution of magnesium nitrate was prepared by dissolving 25.0 g of Mg(NO3)2 in exactly 500. g of water. What is the expected vapor pressure of this solution at 80 oC?

(Vapor pressure of water at 80 oC is 355 torr)

A) 349 torr

B) 372 torr

C) 355 torr

D) 16.9 torr

E) 364 torr

i know the answer but how ?

Explanation / Answer

MOLES OF MG[NO3]2 = 25/148.3 = 0.168 MOLES

SO MOLES OF WATER = 500/18 = 27.77 MOLES

MOLE FRACTION OG MG[NO3]2 = 6.01*10^-3

SO ACCORDING TO RELATIVE LOWERING BY COLLIGATIVE PROPERTIES

P*-P/P* = MOLE FRACTION OF MG[NO3]2

= 23.8-P/23.8 = 6.01*10^-3

= P= 23.65 TORR

VAPOUR PRESSURE OF  MG[NO3]2 AT 25* IS 23.65 TORR

SO NOW AT 80 *C VAPOUR PRESSURE OF VAPOUR = 355 TORR

P* NOW BECOME = 355 TORR

AND THE MOLEFRACTION OF  MG[NO3]2 REMAINS THE SAME AS = 6.01*10^-3

SO

AGAIN APPLYING

P*-P/P* = 6.01*10^-3

= 355-P/355 = 6.01*10^-3

= P= 352.86 TORR

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