$32 g r a m s$ of $S O_{2}$ gas at $S T P$
Which of the following is correct concerning the above statement?

6.02 \times 10^{23} m o l e c u l e s

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11.2 l i t r e s

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58.5 g r a m s / m o l e

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2.0 m o l e

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5 a t o m s

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Solution

The correct option is B $11.2 l i t r e s$
Answer(B)
Moles in 32 grams of $S O 2$ $=$ 32/64 $=$ 0.5 moles
1 mole $=$ 22.4 litres so 0.5 mole $=$ 11.2 litres.

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