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Byju's Answer

Standard XII

Chemistry

Equivalent Mass

10 mL of a so...

Question

$10$ mL of a solution of $K C l$ containing $N a C l$ on evaporation gave $0.93$ g of the mixed salt, which gave $1.865$ g of $A g C l$ by the reaction with $A g N O_{3}$ . The quantity of $N a C l$ in $10$ mL of the solution is (write it in multiples of $100$ ) :

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Solution

$N a C l A g N O_{3} - --- \to A g C l$
$a$ g
$K C l A g N O_{3} - --- \to A g C l$
$b$ g
$∴ a + b = 0.93 . . . (i)$

$∵$ $58.5$ g $N a C l$ gives $143.5$ g $A g C l$ .

$∴$ $a$ g $N a C l$ gives $\frac{143.5 \times a}{58.5}$ g $A g C l$

Similarly,
$∵$ $74.5$ g $K C l$ gives $143.5$ g $A g C l$

$∴$ $b$ g $K C l$ gives $\frac{143.5 \times b}{74.5}$ g $A g C l$

$∴ \frac{143.5 a}{58.5} + \frac{143.5 b}{74.5} = 1.865 . . . (i i)$

Solving Eqs. $(i)$ and $(i i)$ , we get
$a = 0.14$ g
$b = 0.79$ g

So, the answer is $14$ .

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10 mL of a solution of KCl containing NaCl on evaporation gave 0.93 g of the mixed salt, which gave 1.865 g of AgCl by the reaction with AgNO3. The quantity of NaCl in 10 mL of the solution is (write it in multiples of 100) :

$10$ mL of a solution of $K C l$ containing $N a C l$ on evaporation gave $0.93$ g of the mixed salt, which gave $1.865$ g of $A g C l$ by the reaction with $A g N O_{3}$ . The quantity of $N a C l$ in $10$ mL of the solution is (write it in multiples of $100$ ) :