Question

What is the mass in grams of 9.76 × 1012 atoms of naturally occurring beryllium?

Answer 1

The atomic mass of beryllium (Be) is 9 g/mole

Now, 1 mole of any substance contains avogadro's number of that substance. Therefore:

1 mole of Be contains 6.023  10^23 atoms of Be

In other words:

9 grams of Be contains 6.023*10^23 atoms of Be

Therefore, the mass corresponding to 9.76 * 10^12 atoms of Be is:

= 9 g * 9.76 10^12 atoms/6.023*10^23 atoms

= 1.458 * 10^-10 g (or) 1.46 * 10^-10 g

Answer 2

Answer: The mass of given number of atoms of beryllium is $1.46\times 10^{-10}g$

Explanation:

Beryllium is the 4th element of the periodic table.

We know that:

Molar mass of beryllium = 9 g/mol

We are given:

Number of beryllium atoms = $9.76\times 10^{12}$

According to mole concept:

$6.022\times 10^{23}$ number of molecules occupy 1 mole of a gas.

As, $6.022\times 10^{23}$ number of beryllium atoms has a mass of 9 grams.

So, $9.76\times 10^{12}$ atoms of beryllium will have a mass of = $\frac{9g}{6.022\times 10^{23}}\times 9.76\times 10^{12}=1.46\times 10^{-10}g$

Hence, the mass of given number of atoms of beryllium is $1.46\times 10^{-10}g$