Question #474d7

Oct 25, 2015

$1 \text{g}$ helium contains the most particles.

Explanation:

The relative atomic mass in grams of an element contains the number of particles equal to the Avogadro Constant $= 6.02 \times {10}^{23} {\text{mol}}^{- 1}$

So $\frac{m}{A} _ r$ $\times 6.02 \times {10}^{23}$ will give you the number of particles, in this case atoms, which are present.

Using approx. ${A}_{r}$ values:

For $A l$ $n = \frac{1}{27} \times 6.02 \times {10}^{23} = 2.23 \times {10}^{22}$

For $C$ $n = \frac{1}{12} \times 6.02 \times {10}^{23} = 5 \times {10}^{22}$

For $H e$ $n = \frac{1}{4} \times 6.02 \times {10}^{23} = 15 \times {10}^{22}$

For $F e$ $n = \frac{1}{56} \times 6.02 \times {10}^{23} = 1.07 \times {10}^{22}$

So you can see from these data that 1g of He contains the most atoms.