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Standard XII

Chemistry

Percentage Composition

An impure sam...

Question

An impure sample of sodium chloride that weighed $0.50$ g gave $0.90$ g of silver chloride as precipitate on treatment with excess of silver nitrate solution. Calculate the percentage purity of the sample:

72 %

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78 %

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80 %

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None of the above

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Solution

The correct option is A $72 %$
The balanced reaction is as follows:

$N a C l + A g N O_{3} \to A g C l + N a N O_{3}$

$143.5$ g of $A g C l$ is obtained from $58.5$ g of pure $N a C l$ .

$0.9$ g of $A g C l$ is obtained from $\frac{58.5 \times 0.9}{143.5}$ g $= 0.36$ g of pure $N a C l$ .

Percentage purity $= \frac{0.36 \times 100}{0.5} = 72 %$

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An impure sample of sodium chloride that weighed 0.50 g gave 0.90 g of silver chloride as precipitate on treatment with excess of silver nitrate solution. Calculate the percentage purity of the sample:

The correct option is A 72%The balanced reaction is as follows:NaCl+AgNO3→AgCl+NaNO3143.5 g of AgCl is obtained from 58.5 g of pure NaCl.0.9 g of AgCl is obtained from 58.5×0.9143.5g =0.36 g of pure NaCl.Percentage purity =0.36×1000.5=72%

An impure sample of sodium chloride that weighed $0.50$ g gave $0.90$ g of silver chloride as precipitate on treatment with excess of silver nitrate solution. Calculate the percentage purity of the sample: