Which of the following figures does not represent $1 m o l e$ of oxygen gas at $S T P$ ?

11.2 l i t r e s

of gas

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6.022 \times 10^{23}

dioxygen molecules

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22.4 l i t r e s

of gas

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16 g r a m s

of gas

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Solution

The correct option is D $16 g r a m s$ of gas
$Conditions followed by an ideal gas at S T P (a t m)$

$S T P (a t m)$ stands for Standard Temperature and Pressure. At $S T P (a t m)$ , a system is said to have a temperature of $0^{\circ} C$ and the pressure equal to one atmosphere.

$Analyzing the options$

Option $(A)$ :

No. of moles of $O_{2}$ gas:

$= \frac{M a s s o f O_{2} g a s (g)}{M o l a r m a s s o f O_{2} (g m o l^{- 1})}$

Given;

No. of moles of $O_{2}$ gas at STP must be $= 1 m o l$
Molar mass of $O_{2} = 32 g m o l^{- 1}$
Given mass of $O_{2} = 16 g$
$\Rightarrow N o . o f m o l = \frac{M a s s o f O_{2} g a s}{32 g m o l^{- 1}}$

$\Rightarrow$ No. of mol of $O_{2}$ gas $= 0.5$

So, option $(A)$ is incorrect.

Option $(B)$ :

For an ideal gas $(O_{2})$ ;
$P V = n R T$

At $S T P (a t m)$ ;

$P = 1 a t m$
$T = 0^{\circ} C = 273 K$
$n = 1 m o l$ (given in the question)
$R = 0.0821 a t m L m o l^{- 1} K^{- 1}$

Putting the values in ideal gas equation to find the volume of $1 m o l$ of $O_{2}$ gas;

$1 (a t m) \times V = 1 (m o l) \times 0.0821 (a t m L m o l^{- 1} K^{- 1}) \times 273 (K)$

$\Rightarrow V = 22.42 L \approx 22.4 L$

So, option $(B)$ is correct.

Option $(C)$ :

One mole of any substance: $= 6.023 \times 10^{23}$ entities

Hence, it can be said that;

One mole of $O_{2}$ gas: $= 6.023 \times 10^{23} O_{2}$ (dioxygen) molecules

So, option $(C)$ is correct.

Option $(D)$ :

In option $(B)$ , volume of $1 m o l$ of $O_{2}$ gas is calculated as $22.42 L$ .

So, option $(D)$ is incorrect.

Hence, option $(A)$ and $(D)$ are the correct answers.

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