The vapour pressure of pure perfluoro cyclohexane (C6F12) are given by: Solid C6F12:log10P(mm of Hg)=8.68−1895TLiquid C6F12:log10P(mm of Hg)=7.60−1530TThe molar enthalpy of fusion of solid C6F12 is approximately x kJ, the value of x is

For solidlog P=logA−ΔHsub2.303R1T ΔHsub2.303R=1895For liquidlogP=logA−ΔHvap2.303R.1T ΔHvap2.303R=1530 ΔHfusion=ΔHsub−ΔHvap ΔHfusion=7KJ

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The vapour pressure of pure perfluoro cyclohexane $(C_{6} F_{12})$ are given by:
Solid $C_{6} F_{12} : \log_{10} P (m m o f H g) = 8.68 - \frac{1895}{T}$
Liquid $C_{6} F_{12} : \log_{10} P (m m o f H g) = 7.60 - \frac{1530}{T}$
The molar enthalpy of fusion of solid $C_{6} F_{12}$ is approximately x kJ, the value of x is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 7.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

For solid
$\log P = \log A - \frac{Δ H_{s u b}}{2.303 R} (\frac{1}{T}) \frac{Δ H_{s u b}}{2.303 R} = 1895$

For liquid
$\log P = \log A - \frac{Δ H_{v a p}}{2.303 R} . \frac{1}{T} \frac{Δ H_{v a p}}{2.303 R} = 1530 Δ H_{f u s i o n} = Δ H_{s u b} - Δ H_{v a p} Δ H_{f u s i o n} = 7 K J$