SELECT * FROM question_mgmt as q WHERE id=1574 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)

Question:

An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Answer:

Here,

Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar  (Given)

Vapour pressure of pure water at normal boiling point (p10) = 1.013 bar

Mass of solute, (w2) = 2 g

Mass of solvent (water), (w1) = 100 - 2 = 98 g

Molar mass of solvent (water), (M1) = 18 g mol - 1

According to Raoult's law,

(p10 - p1) / p10    =  (w2 x M1 ) / (M2  x w1 )

(1.013 - 1.004) / 1.013 =  (2 x 18) / (M2 x 98 )

0.009 / 1.013  =   (2 x 18) / (M2 x 98 )

M2  =  (2 x 18 x 1.013) / (0.009 x 98)

M2  = 41.35 g mol - 1

Hence, the molar mass of the solute is 41.35 g mol - 1.


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