At 300k when a solute is added to a solvent its vapour pressure over the mercury reduces from 50 mm to 45 mm the value of mole fraction of solute will be
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According to lowering of Vapour pressure
p0 − p/p0 = X2
Where po − vapour pressure of pure solvent
p − vapour pressure of solute
X2 − mole fraction of the solute
50 − 45/50 = X2
X2 = 5/50 = 1/10 = 0.1
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According to lowering of Vapour pressure
p0 − p/p0 = X2
Where po − vapour pressure of pure solvent
p − vapour pressure of solute
X2 − mole fraction of the solute
50 − 45/50 = X2
X2 = 5/50 = 1/10 = 0.1
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