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Question
At 100°C the vapour pressure of a solution containing 6.5 g a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be:
Options
102°C
100°C
101°C
100.52°C
Solution
101°C
Explanation:
`("P"_"A"^0 - "P"_"A")/("P"_"A"^0) = "x"_"B"`
`(760 - 732)/760 = "x"_"B"`
xB = `7/190`
xB = `"m"/("m" + 1000/"M"_"A")`
`7/190 = "m"/("m" + 1000/18)`
`7 "m" + 7000/18` = 190 m
183 m = `7000/18`
m = `7000/3294`
ΔTb = Kb × m = `0.52 xx 7000/3294` = 1.06
Boiling point = 100 + 1.06 = 101°C
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