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
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
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