Number of atoms present in $12 g m$ of $_{6} C^{12}$ will be:

6

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12

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6.02 \times 10^{23}

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6 \times 6.02 \times 10^{23}

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Solution

The correct option is C $6.02 \times 10^{23}$
$1 m o l e = 6.023 \times 10^{23} a t o m s n o . o f m o l e s = \frac{G i v e n m a s s}{M o l a r m a s s} C_{6}^{12} m o l a r m a s s = 12 g H e n c e, n o . o f a t o m s = 6.023 \times 10^{23} a t o m s$

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Similar questions

The number of atoms present in 4.25 g of $({NH}_{3})$ is approximately.

a) $1 \times 10^{23}$

b) $1.5 \times 10^{23}$

c) $0.25 \times 10^{23}$

d) $6.02 \times 10^{23}$

6.02 \times 10^{23}

6.02 \times 10^{23}

4.25 g r a m s

N H_{3}

\times 10^{23}

C u

\times 10^{23}

C

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6.02 \times 10^{23})

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