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

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Answer: The mass of given number of atoms of beryllium is
1.46* 10^(-10)g

Step-by-step 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* 10^(12)

According to mole concept:


6.022* 10^(23) number of molecules occupy 1 mole of a gas.

As,
6.022* 10^(23) number of beryllium atoms has a mass of 9 grams.

So,
9.76* 10^(12) atoms of beryllium will have a mass of =
(9g)/(6.022* 10^(23))* 9.76* 10^(12)=1.46* 10^(-10)g

Hence, the mass of given number of atoms of beryllium is
1.46* 10^(-10)g

User Beach Boys
by
8.2k points
4 votes

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


User Amit Andharia
by
8.0k points

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