1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A mixture of chlorobenzene and wate (immiscible) boils at 90.3oC at an external pressure of 740.2 mm. The vapour pressure of pure water at 90.3oC is 530.1 mm. Calculate the % composition of distillate :

A
H2O= 35%
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
H2O= 22%
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
H2O= 29%
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
H2O= 71%
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

The correct option is C H2O= 29%At boiling point P′M=740.2mmP′H2O=530.1mm∴P′chlorobenzene=740.2−530.1=210.1mmAlso, P′=PM× mole fraction530.1=740.2× mole fraction of H2O∴ mole fraction of H2O=0.716∴ mole fraction of C6H5Cl=0.284Let a g H2O and b g chlorobenzene be present in distillateor wH2O/18wCB112.5+WH2O18=0.716 ....(i)and wH2O/112.5wCB112.5+WH2O18=0.284 ....(ii)By eq. (i)/(ii)or a18×112.5b=0.7160.284or ab=12.8931.5=0.403 .....(iii)Let total distillate be 100 g∴a+b=100 .....(iv)By eqs. (iii) and (iv) b= 71.27%a= 28.73%

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Liquids in Liquids and Raoult's Law
CHEMISTRY
Watch in App
Explore more
Join BYJU'S Learning Program