Number of atoms in $11.2 L$ of $C O_{2}$ at $N T P$ is:

1.5 N_{A}

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0.5 N_{A}

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3 N_{A}

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\frac{2}{3} N_{A}

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Solution

The correct option is A $1.5 N_{A}$
At $N T P$ condition $11.2$ $L$ means as moles ${C O}_{2}$
$∴$ $3 \times 0.5 \times N_{A}$ atoms will be present
$= 1.5$ $N_{A}$
$∴$ Ans :- Option A.

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Similar questions

Calculate the number of molecules and the number of atoms present in 11.2L of Oxygen at N.T.P.

_{4}

_{2}

_{3}

_{2}

\to

_{2}

_{2}

_{4}

A (g) + B (g) \to C (g)

A

N_{A}

${[\frac{- 1 + \sqrt{(- 3)}}{2}]}^{3 n}$ + ${[\frac{- 1 - \sqrt{(- 3)}}{2}]}^{3 n}$

$P_{4} + 3 N a O H + 3 H_{2} O ⟶ 3 N a (H_{2} P O_{2}) + P H_{3}$ is an example of:

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