How many moles of sucrose should be dissolved in `500gms` of water so as to get a solution which has a difference of `104^(@)C` between boiling point and freezing point. `(K_(f)=1.86 K Kg "mol"^(-1).K_(b)=0.52K Kg "mol"^(-1))`

Options

A. `1.68`
B. `3.36`
C. `8.40`
D. `0.840`

Answer

D

Explanation

Boiling point of solution `=` boiling point `+DeltaT_(b)=100+DeltaT_(b)`
Freezing point of solution `=` Freezing point `-DeltaT_(f)=0-DeltaT_(f)`
Diffrence in temperature (given) `=100+DeltaT_(B)-(-DeltaT_(f))`
`104=100+DeltaT_(b)+DeltaT_(f)=100+"molality"xxK_(b)+"molallity"xxK_(f)`
`=100+"molality"(0.52+1.86)`
`therefore "Molality"=(104-100)/(2.38)=(4)/(2.38)=1.68m`
and molality `=("moles"xx1000)/(W_(gm(solvent))),1.68=("moles"xx1000)/(500)`
`therefore` Moles of solute `=(1.68xx500)/(1000)=0.84` moles.