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Equilibrium Constant from Nernst Equation

A mixture con...

Question

A mixture contains $N a C l$ and an unknown chloride $M C l$ . (a) $1$ g of this is dissolved in water, excess of acidified $A g N O_{3}$ solution is added to it so that $2.567$ g of white ppt. is obtained. (b) $1$ g of original mixture is heated to $300$ K. Some vapours come out which are absorbed in $A g N O_{3}$ (acidified) solution. $1.341$ g of white precipitate is obtained. The molar mass of the unknown chloride is (only integer part) :

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Solution

By statement given in (a)
Mass of $N a C l = a$ g
Mass of $M C l = b$ g
$∴ a + b = 1$
Also, Meq. of $N a C l$ + Meq. of $M C l$ = Meq. of $A g N O_{3}$
= Meq. of $A g C l$
= $\frac{2.567}{143.5} \times 1000$
Meq. of $N a C l$ + Meq. of $M C l$ $= 17.889$
Now, as reported in statement (b),
Mixture is heated to $300$ K. Since, $N a C l$ is stable, the possibility seems for the decomposition of $M C l$ and thus, after heating only $M C l$ forms vapour which react with $A g N O_{3}$ .
Meq. of $M C l$ = Meq. of $A g N O_{3}$ = Meq. of $A g C l$
= $\frac{1.341}{143.5} \times 1000$
Meq. of $M C l$ $= 9.345$
Meq. of $N a C l = 17.889 - 9.345 = 8.544$
$\frac{w}{58.5} \times 1000 = 8.544$
$w_{N a C l} = 0.50$ g
Meq. of $M C l$ in mixture $= 9.345$
$\frac{w}{M / 1} \times 1000 = 9.345$
$\frac{1 - 0.50}{M} \times 1000 = 9.345 (∵ a + b = 1)$
$M = 53.50$
$M C l$ may be $N H_{4} C l$ .
So, answer is $53$ .

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