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Solutions question

2021 · Q116
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Solutions question

2021 · Q116

NEETChemistrySolutionsMCQ+4 / −1
The correct option for the value of vapour pressure of a solution at 45∘^\circ∘C with benzene to octane in molar ratio 3 : 2 is :

[At 45∘^\circ∘C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]
  1. A
    350 mm of Hg
  2. B
    160 mm of Hg
  3. C
    168 mm of Hg
  4. D
    336 mm of Hg
View written solutionFree

Correct answer: D

Given : nC6H6:nC8H18=3:2{n_{{C_6}{H_6}}}:{n_{{C_8}{H_{18}}}} = 3:2nC6​H6​​:nC8​H18​​=3:2

So, χC6H6=35,χC8H18=25{\chi _{{C_6}{H_6}}} = {3 \over 5},{\chi _{{C_8}{H_{18}}}} = {2 \over 5}χC6​H6​​=53​,χC8​H18​​=52​

Total vapour pressure of solution,

ps=pC6H6oχC6H6+pC8H18oχC8H18{p_s} = p_{{C_6}{H_6}}^o{\chi _{{C_6}{H_6}}} + p_{{C_8}{H_{18}}}^o{\chi _{{C_8}{H_{18}}}}ps​=pC6​H6​o​χC6​H6​​+pC8​H18​o​χC8​H18​​

=280×35+420×25 = 280 \times {3 \over 5} + 420 \times {2 \over 5}=280×53​+420×52​

=168+168 = 168 + 168=168+168

=336 = 336=336 mm of Hg

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