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Question

A mixture containing KCl and NaCl was dissolved and total halide was determined by titration with silver nitrate. A sample weighing 0.3250 g required 51 mL of 0.1 N solution. Calculate the percentage of each salt in the sample.

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Solution

In the mixture of KCl and NaCl :

74.5 g of KCl requires 170 g of AgNO3 and

58.5 g of NaCl requires 170 g of AgNO3

And in 51mL of 0.1N solution :
weight of AgNO3 = 0.1×170×511000=0.867g

In solution weighing 0.3250g let us consider x gm KCl and the remaining 0.325x gm is NaCl.

molesNaCl+molesKCl=molesAgNO3

x74.5+0.325x58.5=0.867170

x=0.1240gm=weight of KCl in sample

0.201gm=weight of NaCl in sample

percentage of KCl=0.124×1000.325=38.19%

percentage of NaCl=0.201×1000.325=61.81%


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