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0.1 mole of chromium sulphide will contain (valency of Cr = 3, valency of sulphide = 2)

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0.3 g atom of ${Cr}^{+ 2}$ and 0.2 g atom of $S^{- 2}$

0.1 g atom of ${Cr}^{+ 3}$ and 0.1 g atom of $S^{- 2}$

0.2 g atom of ${Cr}^{+ 3}$ and 0.3 g atom of $S^{- 2}$

0.3 g atom of ${Cr}^{+ 3}$ and 0.3 g atom of $S^{- 2}$

answer is C.

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Detailed Solution

$1 {mole of Cr}_{2} S_{3} contains \to 2 {g atoms of Cr}^{+ 3}$

$\Rightarrow 0.1 {mole of Cr}_{2} S_{3} contains \to 0.1 \times 2 = 0.2 {g atoms of Cr}^{+ 3}$

$1 {mole of Cr}_{2} S_{3} contains \to 3 {g atoms of S}^{- 2}$

$\Rightarrow 0.1 {mole of Cr}_{2} S_{3} contains \to 0.1 \times 3 = 0.3 {g atoms of S}^{- 2}$

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