Question

2.66 g of a mixture of KCl and NaCl gave, 5.74 g of dry silver chloride, on treatment with silver nitrate solution. The molar ratio of NaCl to KCl is. (Molar mass of KCI is 74.5 gmol-1, NaCl is 58.5 gmol-' and AgCl is 143.5 gmol-1)​

Answer 1

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Answer 2

Answer:

The ratio for this is 1.021 : 1.639

Explanation:

Following is the reaction of silver nitrate and sodium chloride:

NaCl       +       $AgNO_{3}$ →  $NaNO_{3}$  + AgCl

1 mol                1 mol    

58.5gmol-1    143.5gmol-1

We can see that 1 mol sodium chloride is completely reacting with only 1 mol of silver nitrate.

Hence if x is the amount of the NaCl that is produced, then it will be as follows:

x gram NaCl = 143.5g / 58.5g × x g  

Similarly, for KCl and the silver nitrate, we get the following reaction:

KCl     +   $AgNO_{3}$ →   $KNO_{3}$  + AgCl  

1 mol      1 mol

74.5gmol-1  143.5gmol-1

Therefore a total of 2.66 g of the KCl and the NaCl mixture was present.

Therefore, if x g is the NaCl then KCl = 2.66g - x g

(2.66 -x) g KCl will be producing = 143.5g/ 74.5g × (2.66 - x) g of silver chloride

A total of 5.74 g of dry silver chloride was produced therefore

2.45 × x + 1.92 (2.66 - x) = 5.74

∴ x = NaCl = 1.021 gmol-1

(2.66 - x) gmol-1  = KCl = (2.66 - 1.021) gmol-1 = 1.639 gmol-1

Molar ratio = NaCl : KCl = 1.021 : 1.639

Hence this is the required ratio.