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 `5%` solution (by mass) of cane sugar in water has freezing point of

1.	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.
Answer» 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`

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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.

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