The vapour pressure of pure liquid (molecular
weight = 50) at 25°C is 640 mm of Hg and the
vapour pressure of a solution of a solute in the
liquid at the same temperature is 600 mm of Hg.
Molality of solution is
(A) 3/4
(B) 3/8
(C)4/3
(D) 4/6
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Answer:
c) 4/3
Explanation:
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Given: Vapour pressure of pure liquid - P° = 640mm of Hg
vapor pressure of a solution of the solute in liquid - P = 600mm of Hg
To Find: Molality of the solution - M =?
Solution:
- Molality = moles of solute (n)/ mass of solvent in kg(w₂)
- n = given mass(m')/molecular mass (M')
- According to Raoult's law:-
- P° - P/P° = moles of solute/moles of solvent
- P° - P/P° =
- P° - P/P° = molality of solution (M) × M'₂ /1000 (to convert to kg)
Applying the above Formula to calculate the molality of solution:-
P° - P/P° = M × M'₂/ 1000
640 - 600/640 = M × 50/1000
40/640 = M × 1/20
1/16 = M × 1/20
M = 20/16
M = 5/4
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