1.

A 0.2m aqueous solution of KCl freezes at -0.68^(@)C .Calculate 'I' and the osmotic pressure at 0^(@)C. Assume the volume of solution to be that of pure H_(2)O and K_(f) for H_(2)O is 1.86

Answer»

Solution :`(DeltaT_(f))_("normal")=K_(f)xxm`…………(Eqn 7)
`=1.86xx0.2`
`=0.372`
We have `i=("observed colligative property")/("normal colligative property")`………..(Eqn 9)
`=(0.68)/(0.372)=1.83`
Again, `i=("observed osmotic pressure")/("normal osmatic pressure")`
`:.` observed osmotic pressure `=ixx` normal osmotic pressure
`=i xx c RT`............(Eqn 6)
`=1.83xx0.2xx0.082xx273`
`=8.2` ATM


Discussion

No Comment Found

Related InterviewSolutions