Question

Given 88.0 grams of Boron, how many atoms of Boron (B) are present?

Answer 1

Answer: 77.189 atoms

Explanation:

Boron =10.811

Total grams of boron - atomic mass of boron

88.0-10.811

=77.189

Answer 2

Answer:

$4.9$×$10^{24}$ atoms.

Explanation:

The number of atoms in a substance can be calculated by multiplying the number of moles in the substance by Avogadro's number ($6.022$×$10^{23}$).

Given:

The number of moles in 88.0 grams of Boron.

Weight of boron is 10.81g

It can be calculated as :

$No\ of \ moles\ =\ \frac{88}{10.81}$

$No\ of\ moles\ =\ 8.14moles$

$The \ number \ of \ atoms\ in\ boron\ is\ =\ 8.14 moles$×$6.022$×$10^{23}$

$=4.9$×$10^{24} \ atoms\ of\ boron$

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