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 Vant’s Hoff factor are 2 and 3 respectively.

Moles of NaCl = x/58.5

Moles of CaCl₂ = 228-x

And hence molality of NaCl and CaCl₂ are 

Molality of NaCl = ×/58.5×10kg

Molality of 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

Mole percentage (%) 0f NaCl = (2/2+1)x100= 66.67%

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