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Question
Calculate the mass of a non-volatile solute (molar mass 40 g mol−1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Solution
Let the vapour pressure of pure octane be `"p"_1^0`.
Then, the vapour pressure of the octane after dissolving the non-volatile solute is `80/100 "p"_1^0 = 0.8 "p"_1^0`
Molar mass of solute, M2 = 40 g mol−1
Mass of octane, w1 = 114 g
Molar mass of octane, (C8H18), M1 = 8 × 12 + 18 × 1
= 114 g mol−1
Applying the relation,
`("p"_1^0 - "p"_1)/("p"_1^0) = ("w"_2xx"M"_1)/("M"_2xx"w"_1)`
`=>("p"_1^0 - 0.8 "p"_1^0)/("p"_1^0)` = `("w"_2xx114)/(40xx114)`
`=> (0.2 "p"_1^0)/"p"_1^0 = "w"_2/40`
`=> 0.2 = "w"_2/40`
`=> "w"_2 = 8 "g"`
Hence, the required mass of the solute is 8 g.
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