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?
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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there is a little mistake in formula right formula is (p1* - p1) / p1 = (w2 x M1 ) / (M2 x w1 ) by this formula the answer is 40.9 which is correct also mention in NCERT question 2.15
Very helpful
In the question,it's only given as a solvent.Why did u take it as water?
Thank you so much very informative
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isme vapour presure water ka to diya hi na h
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