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A certain sample of cuprous sulphide is found to have composition \[C{{u}_{1.8}}S\] because of incorporation of \[C{{u}^{2+}}\] ions in the lattice. What is the mole percentage of the \[C{{u}^{2+}}\] in total copper content in this crystal?
A. \[99.8%\]
B. \[11.11%\]
C. \[88.88%\]
D. None of these

Last updated date: 20th Jun 2024
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Hint: In this type of question, you need to find out the total loss in the given sample \[C{{u}_{1.8}}S\] and then you can find out the mole percentage of the \[C{{u}^{2+}}\] in total copper content in this crystal.

Complete step by step answer:
We know molecular formula of pure compound with \[C{{u}^{+}}\] is \[C{{u}_{2}}S\], but due to impurity of \[C{{u}^{2+}}\] molecular formula becomes \[C{{u}_{1.8}}S\].
We can write this as: $C{{u}_{2}}S\underset{\text{Impurity added}}{\longleftrightarrow}C{{u}_{1.8}}S$
From this we can find total loss copper i.e.,
Therefore, percentage of \[C{{u}^{2+}}\] in total copper content is given as
$=\,\dfrac{Initial\,-\,Final}{Final}\,\,\times 100$
Substituting the values in above formula we get,
$=\dfrac{0.2}{1.8}\times 100=11.11%$
Hence, mole percentage of \[C{{u}^{2+}}\] in total copper content is $11.11%$

So, the correct answer is Option B.

Additional Information:
In both \[C{{u}_{2}}S\] and \[C{{u}_{1.8}}S\], electrical neutrality is maintained. Atom is said to be electrically neutral when the overall charge of the atom is zero.
Copper sulphide is a chemical compound of copper and sulphur, it has the chemical formula \[C{{u}_{2}}S\]. It is basically found in nature as the mineral chalcocite.
Copper sulfide ores must be concentrated before they can be economically transported and smelted.

Note: You should know the chemical formula of copper sulphide so that you can easily find out the loss from the sample which is impure. Also to find the loss of percentage composition of a particular element from the given compound we subtract final composition from initial composition and divide by final. This will give the composition for percentage calculation we have to multiply it with hundred.