SELECT * FROM question_mgmt as q WHERE id=1721 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=57 and ex_no='2' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th chemistry Solutions 100 g of liquid a molar mass 140 g mol 1 was dis

Question:

100 g of liquid A (molar mass 140 g mol-1) was dissolved in 1000 g of liquid B (molar mass 180 g mol-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Answer:

Number of Moles of Liquid A, nA = 100 / 140 = 0.714

Number of Moles of Liquid B, nB = 1000 / 180 = 5.556

Then Mole fraction of A = nA / nA + nB = 0.714 / 0.714 + 5.556

= 0.114

Now Mole of fraction of B = 1 - 0.114 = 0.886

Now ptotal  = pA + pB

Or ptotal = p°AXA + p°BxB

OR

475 = p°A X 0.114 + 500 X 0.886

OR

A = 280.7 torr

Therefore vapour pressure of pure A = 280.7 torr

Vapour pressure of A in solution = 280.7 x 0.114

= 32 torr

Now

p= p°AXA

Or

= pA   /  XA

⇒ 32 / 0.114

= 280.7 torr

Hence, the vapour pressure of pure liquid A is 280.7 torr.


SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='3' AND subId='9' AND ex_no!=0 AND status=1 and id!=1721 ORDER BY views desc, last_viewed_on desc limit 0,10
SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='3' AND subId='9' AND ex_no!=0 AND status=1 and id!=1721 ORDER BY last_viewed_on desc limit 0,10

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