A solution containing 0.1 mole of naphthalene and 0.9 mole of benzene is cooled out until some benzene freezes out. The solution is then decanted from the solid and warmed to 353 K. At 353 K the vapour pressure was found to be 670 torr. The freezing point and normal boiling points of benzene are 278.5 K and 353 K respectively. The enthalpy of fusion = 10.67 kJ/mol
(i)Amount of benzene solidified is –
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i)
Sol. At 353 K ; p solution. = 670 torr ; p solvent = 760 torr
or 
= x 2 or 
= 
= 
= 
or 
= 0.1 + n benz
or n benz = 0.845 – 0.1 = 0.745
∴ moles of benzene freezes out = 0.9 – 0.745
= 0.155 moles
= 0.155 × 78 = 12.09 gm
(ii)
Sol. 
= 
= 0.118
(iii)
Sol. k f = 
= 
k f = 
= 4.714
Δ T f = k f × m = 4.714 × 
×10 3 Δ T f = 
= 15.1
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