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RiverTmasco RiverTmasco
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11 years ago
Would a 1m aqueous solution of glucose, C6H1206, or a 1m aqueous solution of the ionic compound,(NH4)3 PO4, have a lower freezing point? Explain your reasoning and prove it with Tf calculations. Kf for water is 1.86 degrees Celcius/m.

Thank you.
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wrote...
11 years ago
delT = Kf*m*i.

i = 1 for glucose, and i > 1 for ammonium phosphate (depending on how much it dissociates, if it's completely, n = 4 I think). Plugging in a higher value for i will result in a larger delta T, so the freezing point will change more.
wrote...
11 years ago
NH4)3 PO4 since that would dissociate into 4 m.

Tf = i*m*Kf

i = van hoffs constant or how much something dissociates

For NH4)3 PO4
i = 4

since it dissociates into 3 NH4 + n PO4 3-

So 4*1*1.86 = Tf

then subtract 0 - Tb to get freezing point for NH4)3 PO4

For molecular compound i = 1

so Tf = 1*1*1.86 = 1.86

0-1.86 = -1.86

The freezing pt for NH4)3 PO4 should be 4*-1.86
wrote...
11 years ago
lowering of freezing point is a colligative property. Colligative property depends upon no. of particles . so more the no. of particles of solute more the property. Clearly glucose ( as it don't dissociates) will have lesser no. of paticles than 1m ionic compound, so ionic compound will have lower freezing point than 1m glucose.
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