Question

Mole fraction of a non-electrolyte in aqueous solution is `0.07`. If `K_(f)` is `1.86^(@) "mol"^(-1)kg`, depression if `f.p.,DeltaT_(f),` is:
A. `0.26^(@)`
B. `1.86^(@)`
C. `0.13^(@)`
D. `7.78^(@)`

Answer 1

Correct Answer - D
Firstly we have to convert mole fraction into molality.
`"Molality"=(X_("solute"))/(X_("solvent")M_("solvent")//1000)=(0.07xx1000)/(0.93xx18)=4.18`
Now, `DeltaT_(f)=k_(f)m`
`=1.86xx4.18=7.78^(@)`.