What is the freezing point of 0.1% zinc chloride solution, dissociating 80%
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What is the freezing point of 0.1% zinc chloride solution, dissociating 80%

[From: ] [author: ] [Date: 11-05-07] [Hit: ]
alpha = 80%=0.soi= 1+0.8*(3-1)=2.osmolarity = molarit*2.6 =0.00735*2.......
molecular weight of zinc chloride is 136 & freezing point of molar solution of non-electrolyte is -1.86 degree Celsius.

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First we calculate the number of moles of Zn Cl2 in order to have the molarity

0.1% = 0.001 correspond in weight to 1g of ZnCl2 by liter of water which weighs 1kg =1000g

so the molarity of ZnCl2 is 1/136=0.00735M

now , we must calculate the osmolarity
ZnCl2 being a weak electrolyte you use formula osmolarity = molarity* i
1=1+alpha *(p-1)
alpha = 80%=0.8
p is number of ions given by the dissociation of 1 molecule of ZnCl2 = 1+2=3
so i= 1+0.8*(3-1)=2.6
osmolarity = molarit*2.6 =0.00735*2.6=0.0191

And depression of freezing point is delta t = -1.86*osmolarity= -1.86*0.0191=-0.036C

freezing point is -0.036C

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The equation to be used is:
1000 x Kf x w x (1+ 2x)
^ T = --------------------------------------
W x M
Kf =1.86 ; w =0.1 g ; W= 100 g
M = 136 ; x = 0.8 as ZnCl2 dissociates to give 3 particles total particles in solution are 1+2(0.8)
Hence ^T =0.01368 and freezing point will be - 0.01368 C
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keywords: chloride,of,solution,0.1%,dissociating,is,zinc,point,freezing,80%,What,the,What is the freezing point of 0.1% zinc chloride solution, dissociating 80%
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