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A `5%` solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose (by mass) in water. The freezing point of pure water is 273.15 K.
A. 271 K
B. 273.15 K
C. 269.07 K
D. 277.23 K

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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`]

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