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Q.

The number of chlorine atoms in 20ml of chlorine gas at STP is___ $x 10^{21} (R = 0.0832 B r) N_{A} = 6.023 x 10^{23}$

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answer is 1.06.

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Detailed Solution

$P V = n R T$

$n = \frac{1 \times 20 \times 10^{- 3}}{0 .0821 \times 273} = 0 .89 \times 10^{- 3}$

$1 m o l e____2 \times 6 \times 10^{23} Cl atoms$

$10^{- 3} \times 0.89_____?$

$= 1.06 \times 10^{21}$

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