A mixture contains 1 mole of volati...
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A mixture contains 1 mole of volatile liquid A (P =100 mm Hg) and 3 moles of volatille liquid BP: 80 mm Hg). If solution behaves ideally, the total vapour pressure of the distillate is (A) 85 mm Hg (B) 85.88 mm Hg (C)90 mm Hg (D) 92 mm Hg

JEE/Engineering Exams
Chemistry
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sulle muthre behaces cdealey we wul we Raolts law Moles of ( _{n}^{text {hand }}=n_{B}=frac{text { moles of } B}{text { Total moles }}=frac{3}{4} ) Then, ( quad P_{A}=n_{A} times P^{0} A ) [ begin{array}{c} P_{A}=frac{1}{9} times 100 mathrm{mm} mathrm{Hg}=25 mathrm{mm} mathrm{hg} P_{B}=mathrm{MB} times P_{B}=frac{3}{4} times 80 mathrm{mm} mathrm{Mg} =60 mathrm{mm} mathrm{Mg} end{array} ] The pal parmule ( Rightarrow quad P_{7}=P_{A}+P_{B}=25+60 ) ( =85 mathrm{mm} mathrm{Hg} )
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