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NCERT Class XII Chemistry
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Question : 20 of 53
Marks: +1, -0
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose in water if freezing point of pure water is 273.15 K.
Solution:  
Molality of sugar solution =5342×1000100=0.146 m= \frac{5}{342} \times \frac{1000}{100} = 0.146 \text{ m}
ΔTf\Delta T_{f} for sugar solution =273.15−271=2.15=273.15-271=2.15
ΔTf=Kf×m    ∴Kf=ΔTfm=2.150.146\Delta T_{f}=K_{f} \times m \;\; \therefore K_{f}=\frac{\Delta T_{f}}{m}= \frac{2.15}{0.146}
Molality of glucose solution =5180×1000100=0.278 m= \frac{5}{180} \times \frac{1000}{100} = 0.278 \text{ m}
∴    ΔTf\therefore \;\; \Delta T_{f} (Glucose) =2.150.146×0.278=4.09∘= \frac{2.15}{0.146} \times 0.278 = 4.09^{\circ}
∴    \therefore \;\; Freezing point of glucose solution =273.15−4.09=269.06 K=273.15-4.09=269.06 \text{ K}
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