Welcome to ChemZipper !!!: A mixture of weighing 228 g contain CaCl2 and NaCl . If this mixture is dissolved in 10 kg of water and form ideal solution that boil at 100.364 ℃ The mol % of NaCl in mixture is [ kb of water = 0.52 K per mol Kg]

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A mixture of weighing 228 g contain CaCl2 and NaCl . If this mixture is dissolved in 10 kg of water and form ideal solution that boil at 100.364 ℃ The mol % of NaCl in mixture is [ kb of water = 0.52 K per mol Kg]

Given that 228 g mixture of NaCl and CaCl2 consider that x g NaCl and (228-x)gm  CaCl2 present in this mixture:

NaCl and CaCl2 both are strong electrolytes hence bant Hoff facter are 2 and 3 respectively.

Moles of NaCl = x/ 58.5
Moles of CaCl2 = 228-x
And hence molality of NaCl and CaCl2 are 
m NaCl = ×/58.5×10kg
m CaCl2= 228-x/111×10kg

We know that 
∆Tb = i1m1+i2m2
100.364= 2(×/58.5×10kg)+3(228-x/111×10kg)

On solving we got x= 117 g 

Moles of NaCl= 117/58.5 =2
Moles of CaCl2= 111/111=1

Mol% 0f NaCl = 66.67%

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