The vapour pressure of a pure liquid is 0.80 atm. When a non-volatile solute is added to this liquid, its vapour pressure drops to 0.60 atm. The mole
The vapour pressure of a pure liquid is 0.80 atm. When a non-volatile solute is added to this liquid, its vapour pressure drops to 0.60 atm. The mole fraction of the solute in the solution is
A. 0.75
B. 0.2
C. 0.25
D. 0.85
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Correct Answer - C
`(p^(@)-p_(s))/(p^(@)) = x_("solute")`
`:. x_("solute") = (80-60)/(80) = (1)/(4) = 0.25`
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