19.7 kg of gold was recovered from a smuggler. The atoms of gold recovered are:

100

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6.02 \times 10^{23}

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6.02 \times 10^{24}

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6.02 \times 10^{25}

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Solution

The correct option is C $6.02 \times 10^{25}$
The mass of gold $= 19.7$ kg $= 19.7 \times 1000$ g $= 19700$ g.

The atomic mass of gold is 197 g/mol.

The number of moles of gold $\frac{19700 g}{197 g / m o l} = 100$ mol.

The number of atoms of gold $= 100 \times 6.02 \times 10^{23} = 6.02 \times 10^{25}$ .

So, the correct option is $D$

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6.02 \times 10^{23}

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Four grams of caustic soda contains :

(a) $6.02 \times 10^{23} a t o m s o f i t,$
(b) 4 g atom of sodium,
(c) $6.02 \times 10^{22}$ molecules

(d) 4 moles of NaOH.

6.02 \times 10^{23}

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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) $6.02 \times 10^{23}$

A beaker contains $6.02 \times 10^{24}$ molecules of water . How many mole of atoms is this?

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