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Find the freezing point of solution containing 3.6 g of glucose dissolved in 50g ofwater kg of water |
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Answer» The molecular weight of glucose C
H 12
O 6
=6(12)+12(1)+6(16)=180 g/mol. The number of moles of glucose = 180g/mol 0.625g
=0.00347moles Mass of WATER =102.8g=0.1028kg The molality of glucose m= 0.1028kg 0.00347moles
=0.0338mol/kg The depression in the freezing point ΔT f
=K f
m=1.87Kkg/mol×0.0338mol/kg=0.0632K Freezing point of water =273 K The freezing point of the SOLUTION =273+0.0632=273.0632K |
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