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The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following element contains the greatest number of atoms?

#### Options

\[\ce{4 g He}\]

\[\ce{46 g Na}\]

\[\ce{0.40 g Ca}\]

\[\ce{12 g He}\]

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#### Solution

**\[\ce{12 g He}\]**

**Explanation:**

**(A)** The number of moles is given by the following formula,

Moles = `"Mass"/"Molar mass"` .......(1)

The number of moles of He is calculated by using equation (1) as follows,

Moles of O_{2} = `( 4 g)/((4 g)/(mol)` = 1 mol

The number of atoms can be calculated as, number of moles = `"Numbers of atoms"/"Avogadro's number"` .......(2)

On substituting the values in the above equation:

1 mol = `"Number of atoms"/(6.022 xx 100^23)`

Number of atoms = `1 xx 6.022 xx 10^23`

**(B)** The number of moles of Na is calculated by using equation (1) as follows,

Moles of Na = `(46 g)/((23 g)/(mol)` = 2 mol

The number of atoms can be calculated by using equation (2) as follows,

2 mol = `"Number of atoms"/(6.022 xx 10^23)`

Number of atoms = `2 xx 6.022 xx 10^23`

**(C)** The number of moles Ca is calculated by using equation (1) as follows,

Moles of Ca = `(0.40 g)/((40 g)/(mol)` = 0.01 mol

The number of atoms can be calculated by using equation (2) as follows,

0.01 mol = `"Number of atoms"/(6.022 xx 10^23)`

Number of atoms = `0.01 xx 6.022 xx 10^23`

**(D)** The number of moles of He is calculated by using equation (1) as follows,

Moles of He = `(12 g)/((4 g)/(mol)` = 3 mol

The number of atoms can be calculated by using equation (2) as follows,

3 mil = `"Number of atoms"/(6.022 xx 10^23)`

Number of atoms = `3 xx 6.022 xx 10^23`

Thus, 12 g He contains the highest number of atoms.

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