Correct Answer - C
`DeltaT=K_(f)xx(w_(B)xx1000)/(m_(B)xxw_(A))`
`2.15=K_(f)xx(5xx1000)/(342xx95)` ....(i) (for sucorose)
`DeltaT=K_(f)xx(5xx1000)/(180xx95)`....(ii) (for glucose)
Dividing eq. (i) by eq. (ii) we get,
`DeltaT=4.18K`
`T=T_(0)-4.18`
`= 273.15-4.18=269.07K`]