Correct Answer - `269.07 K`

`W_(2)=5g,W_(solution)=100g, Mw_(2)=342g`

`(C_(12)H_(22)O_(11))`

First case

`W_(1)=100-5=95g`

`DeltaT_(f)=273.15-271=2.15K`

`DeltaT_(f)=K_(f)xx(W_(2)xx1000)/(Mw_(2)xxW_(1))`

`2.15=K_(f)xx(5xx1000)/(342xx95)`

Solve for`K_(f)impliesK_(f)=13.97`

Second case

`W_(2)=5g, W_(1)=95g, ` ltbr. `Mw_(2)` of glucose `(C_(6)H_(12)O_(6))=180g`

`DeltaT_(g)=13.97xx(5xx1000)/(180xx95)=4.085`

`273.15-T_(2)=4.085`

`T_(2)=273.15-4.085=269.07K`