Volume occupied by 1 gram of oxygen at $N T P$ is:

0.7

L

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2.4

L

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2.24

L

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5.2

L

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Solution

The correct option is A 0.7 $L$
At NTP, $1$ mole of oxygen occupies $22.4 L$ .
$1$ mole of oxygen is equal to $32 g$
Hence 1 g will occupy $\frac{1}{32} \times 22.4 = 0.7 L$

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Volume occupied by 1 gram of oxygen at NTP is:

The correct option is A 0.7 LAt NTP, 1 mole of oxygen occupies 22.4 L.1 mole of oxygen is equal to 32 gHence 1 g will occupy 132×22.4=0.7L

Volume occupied by 1 gram of oxygen at $N T P$ is:

The correct option is A 0.7 $L$
At NTP, $1$ mole of oxygen occupies $22.4 L$ .
$1$ mole of oxygen is equal to $32 g$
Hence 1 g will occupy $\frac{1}{32} \times 22.4 = 0.7 L$