Chapter U3.16, Problem 5E

Interpretation Introduction

Interpretation: Whether 8.0 g of He or 40.0 g of argon gas more atoms needs to be determined.

Concept Introduction:

In 1 mol of any substance there are $6.022\times {10}^{23}$ units of substance are present. This is known as Avogadro’s number and represented by symbol $N_A$ .

Answer to Problem 5E

, 8.0 g of He gas has more number of atoms than 40 g of Ar gas.

Explanation of Solution

The mass of helium is 8.0 g and that of argon is 40.0 g. To calculate the number of atoms, the number of moles of each gas needs to be calculated first.

The number of moles can be calculated using the following formula:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

The molar mass of He and Ar gas is 4 g/mol and 40 g/mol respectively. Thus, number of moles can be calculated as follows:

  $n_{He}=\frac{8.0\text{ g}}{4\text{ g/mol}}=2.0\text{ mol}$

Since, in 1 mol there are $6.022\times {10}^{23}$ atoms thus, the number of atoms of He will be:

  $2\times 6.022\times {10}^{23}=12.044\times {10}^{23}$ .

Similarly,

  $n_{Ar}=\frac{40.0\text{ g}}{40\text{ g/mol}}=1.0\text{ mol}$

Since, in 1 mol there are $6.022\times {10}^{23}$ atoms thus, the number of atoms of Ar will be $6.022\times {10}^{23}$ .

Conclusion

Therefore, 8.0 g of He gas has more number of atoms than 40 g of Ar gas.