Question

1. approximately how much would 8.0 × 10²⁴ atoms of silver weigh, in grams? (see example 4 for help with this problem.)

Explanation:

Step1: Recall Avogadro's number

Avogadro's number $N_A = 6.022\times 10^{23}\text{ atoms/mol}$.

Step2: Calculate the number of moles of silver

The number of moles $n$ of silver is given by $n=\frac{N}{N_A}$, where $N = 8.0\times 10^{24}$ atoms. So $n=\frac{8.0\times 10^{24}\text{ atoms}}{6.022\times 10^{23}\text{ atoms/mol}}\approx13.3\text{ mol}$.

Step3: Recall the molar - mass of silver

The molar - mass of silver $M$ is $M = 107.87\text{ g/mol}$.

Step4: Calculate the mass of silver

The mass $m$ of silver is $m=n\times M$. Substituting $n = 13.3\text{ mol}$ and $M = 107.87\text{ g/mol}$, we get $m=13.3\text{ mol}\times107.87\text{ g/mol}\approx1435\text{ g}$.

Answer:

$1435$ g