The number of atoms present in $92$ amu of $N a$ is:

92

atoms

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\frac{N_{A}}{92} \times 23

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\frac{N_{A}}{23} \times 92

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4

atoms

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Solution

The correct option is D $\frac{N_{A}}{23} \times 92$
Moles of $N a$ = $\frac{92}{23} = 4$ moles

$1$ mole of $N a$ contains $6.022 \times 10^{23}$ atoms
$∴$ $4$ mole contains $4 \times 6.022 \times 10^{23}$
i.e. $\frac{92}{23} \times N_{A}$ atoms

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The number of atoms present in 92 amu of Na is:

The correct option is D NA23×92Moles of Na= 9223=4 moles1 mole of Na contains 6.022×1023 atoms∴ 4 mole contains 4×6.022×1023i.e. 9223×NA atoms

The number of atoms present in $92$ amu of $N a$ is:

The correct option is D $\frac{N_{A}}{23} \times 92$
Moles of $N a$ = $\frac{92}{23} = 4$ moles

$1$ mole of $N a$ contains $6.022 \times 10^{23}$ atoms
$∴$ $4$ mole contains $4 \times 6.022 \times 10^{23}$
i.e. $\frac{92}{23} \times N_{A}$ atoms