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The percentage of available chlorine in a sample of bleaching powder, $CaOC{l_2}.2{H_2}O$,is:
(A) 30
(B) 50
(C) 43.5
(D) 59.9

Answer
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Hint: To solve this question, you must know the atomic masses of all the elements given in the molecular formula above. A prior knowledge of conservation of mass and basic mathematics is also essential.

Complete answer:
> Bleaching powder is a white powder, which on contact with water decomposes. During decomposition its odor is like that of gaseous chlorine. It is used for bleaching clothes, as a bleaching agent for cotton, linen, etc. and is relied heavily on in the textile industry.
> Molecular Formula of bleaching powder: $CaOC{l_2}.2{H_2}O$
We know, atomic mass of:
$Ca = 40u$
$O = 16u$
\[Cl = 35.5u\]
$H = 1u$
So atomic mass of individual elements in bleaching powder $(CaOC{l_2}.2{H_2}O)$ is:
$Ca \times 1 = 40 \times 1 = 40u \\O \times 3 = 16 \times 3 = 48u \\Cl \times 2 = 35.5 \times 2 = 71u \\
H \times 4 = 1 \times 4 = 4u \\ $
> Total molecular mass of bleaching powder $(CaOC{l_2}.2{H_2}O) = (40 + 32 + 71 + 4)u = 147u$
> To calculate the percentage of an element present in a molecule, we have to divide the atomic mass of the element with the molecular mass of the molecule and multiply it by hundred.
> The maximum percentage of $Cl$ in a sample of bleaching powder $(CaOC{l_2}.2{H_2}O)$
$\dfrac{{71}}{{147}} \times 100 = 43.5\% $
> The maximum percentage of Chlorine$(Cl)$ in a sample of bleaching powder $(CaOC{l_2}.2{H_2}O)$ is: $43.5\% $
Remember, this is just in theory, in reality the percentage will be less, as even in a highest purified substance some level of impurity is always present. This can be further improved as technology and purifying methods improve.

Note: The main Ingredient for in bleaching powder $(CaOC{l_2}.2{H_2}O)$ is an inorganic compound calcium hypochlorite$(Ca{(ClO)_2})$. Even though bleaching powder is white in color, the commercial samples appear yellow due to some level of exposure to air, which can lead to slow decomposition.