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7 Mar 2008 19:33:15 IST

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239

dep in freezing point

None

the molecular wts of NaCl & glucose are determined using the dep of freezing point mtd

compared to their theoretical wts what will their observed molecular wts bewhen determined by above mtd ??

pls help asap

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Akhil

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7 Mar 2008 19:37:05 IST

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NaCl is ionic salt...almost decomposes completely...
so the num of particles doubles nearly...so depression in freezing point doubles
hence calculated molality is double of actual.
and we get mol wt as half of original...
in glucose i think it remains same as i=1

LAMPARD

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7 Mar 2008 19:39:49 IST

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When dep in freezing pt. method is used,a factor called vant hoff factor comeds in2 picture..for NaCl,its 2 and for glucose its 1...its basically the no. of particles the molecule will dissciate in2..so,by dat method,mol. wt. of NaCl will be half while dat of glucose will remain same...
More theory on vant hoff factor..
In NaCl for example,

i = 2

indicates that by using the colligative properties equations without the

i

factor, we're actually calculating the molar mass of each of the Na⁺ and Cl^- ions in solution. The molar mass of NaCl should be twice that number because it originally dissociated to produce one Na⁺ ion and one Cl^- ion per formula unit of NaCl; the sum of the masses of the two ions should yield the mass of one formula unit of NaCl. Note that even if the mass of Na⁺ and Cl^- weren't equal (which is the case), we're only calculating the average mass of the two particles using our colligative properties analysis. Hence multiplying that average number by two, we would be calculating the molar mass of NaCl.

ankur khurana

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7 Mar 2008 19:41:39 IST

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lampard has given right answer

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