Question

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

• \[\ce{4 g He}\]

• \[\ce{46 g Na}\]

• \[\ce{0.40 g Ca}\]

• \[\ce{12 g He}\]

Answer 1

\[\ce{12 g He}\]

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 O₂ = `( 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.