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Molecular Mass

A sample of m...

Question

A sample of mixture of $C a C l_{2}$ and $N a C l$ weighing $5.55$ g was treated to precipitate all the $C a$ as $C a C O_{3}$ which was then heated and quantitatively converted to $1.68$ g of $C a O$ . The mass percentage of $C a C l_{2}$ in the mixture is;

A

60

%

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B

70

%

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C

45

%

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D

56

%

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Solution

The correct option is D $60$ %
$C a$ present in $1.68$ g $C a O$ is $1.2$ g.
So, original mixture has $1.2$ g $C a$ means $1.2 \times \frac{111}{40} = 3.33$ g $C a C l_{2}$ .
Hence, mass percent of $C a C l_{2}$ $= \frac{3.33}{5.55} \times 100 = 60$ %.

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