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Standard XII

Chemistry

Mole Concept

A mixture of ...

Question

A mixture of $K B r$ and $N a B r$ weighing 0.560 g was treated with aq. $A g^{+}$ ion and all bromide ion was recovered as 0.970 g of pure $A g B r .$ What was the fraction by weight of $K B r$ in the sample?

0.1332

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0.2378

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0.4378

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0.3478

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Solution

The correct option is B $0.2378$
$K B r + N a B r A g^{+} - - \to A g B r$

Let $x$ g be the weight of $K B r$ in the sample.
Then ( $0.560 - x$ ) g be the weight of $N a B r$ .
Because no. of moles of $B r$ atoms are conserved in the reaction, applying POAC on $B r$ atom,
Moles of $B r$ in $K B r$ + moles of $B r$ in $N a B r$ = moles of $B r$ in $A g B r$
$\Rightarrow \frac{x}{119} + \frac{0.560 - x}{103} = \frac{0.97}{188}$

$\Rightarrow x = 0.1332$ g
Fraction of $K B r$ in mixture $= \frac{0.1332}{0.560} = 0.2378$

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Mole Concept

Standard XII Chemistry

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A mixture of KBr and NaBr weighing 0.560 g was treated with aq. Ag+ ion and all bromide ion was recovered as 0.970 g of pure AgBr. What was the fraction by weight of KBr in the sample?

A mixture of $K B r$ and $N a B r$ weighing 0.560 g was treated with aq. $A g^{+}$ ion and all bromide ion was recovered as 0.970 g of pure $A g B r .$ What was the fraction by weight of $K B r$ in the sample?