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Chemistry - Liquid Solution Question with Solution | TestHub

ChemistryLiquid SolutionRaoult's LawMedium2 minQB

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For a liquid normal boiling points is -173°C then at 2 atm pressure it's boiling point should be nearly $(Δ H_{vap}$ . $= 200 cal / mole, R = 2 cal / mol - K, ln 2 = 0.7)$

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Answer:

Solution:

$ln \frac{P _{2}}{P _{1}} = \frac{Δ H}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$

$ln \frac{2}{1} = \frac{200}{2} [\frac{1}{100} - \frac{1}{T _{2}}]$

$T_{2} = \frac{100}{0.3} = 333.33 K = 60.3 3^{\circ} C$

Stream:JEESubject:ChemistryTopic:Liquid SolutionSubtopic:Raoult's Law

⏱ 2mℹ️ Source: QB

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