Calculate the vapor pressure of a solution containing 9 g of glucose in 162 g of water at 293 k
Answers
The vapor pressure of the solution will be 17.45 mm of Hg
Explanation :
Given,
weight of water = 162 g
=> number of moles of water, n₁ = 162/18 = 19
weight of glucose = 9 gm
=> number of moles of glucose, n₂ = 9/180 = 1/20
we know that vapor pressure of pure water at 293 k, P⁰ = 17.5 mm of Hg
From Raoult's law,
(P⁰-P)/P⁰ = n₂/n₁
=> 1 - P/P⁰ = 1/(20x19)
=> P/P⁰ = 379/380 = 0.99
=> P = 17.5 x 0.99 = 17.45 mm of Hg
Hence the vapor pressure of the solution will be 17.45 mm of Hg
Answer:
Explanation:
Given:
Weight of water=165gm
Weight of glucose=9gm
=> no. of moles of water,n1=162/18=19;
no. of moles of glucose,n2=9/180=1/20
We know that vapour pressure of pure water at 294k is
P°=17.5mm of Hg
From raoult's law
(P°-P/P°)=(n1/n2)
=>1-P/P°=1/20×10
=>
P/P°=378/380=0.99
=> P=17.5×0.99=17.45mmof Hg
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