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  • Calculate the freezing point of a solution when 3 g of CaCI2(M = 111 g mol-1) was dissolved in 100 g of water, assuming CaCI2  undergoes complete ionization.(Kf for water = 1.86 K kg mol-1)</p

Calculate the freezing point of a solution when 3 g of CaCI2(M = 111 g mol-1) was dissolved in 100 g of water, assuming CaCI2  undergoes complete ionization.(Kf for water = 1.86 K kg mol-1)

 

 

 

 
 
 
 
 

Answers (1)

Tf = ?        Kf (water) = 1086 K kg mol-1
Mass of CaCI2 = 3g = 0.003 kg
Mass of water = 100g = 0.1 kg
MB = 111 g mol-1 = 0.111 kg mol-1
complete ionisation , i = 3
\left [ CaCI_{2}\rightarrow Ca^{2+} +2CI^{-}\right ]
\Delta T_{f}= i\times k_{f}\times m
\Delta T_{f}= \frac{3\times 1\cdot 86\times 0\cdot 003}{0\cdot 111\times 0\cdot 1}
\Delta T_{f}= 1\cdot 50k
T_{f}= T^{0}_{f}-\Delta T_{f}
T_{f}= 273k-1\cdot 50k
T_{f}= 271\cdot 5k

Posted by

Sumit Saini

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