In rhombic sulphur, S-atoms are joined into $S_{8}$ molecules with density $2.07 g m / c m^{3}$ .

Number of $S_{8}$ molecules in a crystal of 0.568 $m m^{3}$ is

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$1.336 \times 10^{21}$

$2.765 \times 10^{18}$

$2.765 \times 10^{21}$

$2.212 \times 10^{22}$

answer is D.

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Detailed Solution

Ans-4
SOL- $m a s s = 0.568 \times 10^{- 3} \times 2.07 g$
$m o l e o f S_{8} = \frac{0.568 \times 10^{- 3} \times 2.07}{256} c m^{3} m o l e c u l e s = 4.59 \times 10^{- 6} \times 6.02 \times 10^{23} \Rightarrow 2.765 \times 10^{18}$

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