Question

How many atoms of Boron are present in 35.76 g of Boron?

Answer 1

Molar mass of `B` `=` `10.81` `g*mol^-1`. If there are 36 odd grams of `B`, there are approx. `7/2` `xx` `N_A` boron atoms, where `N_A` is Avogadro's number.

Explanation:

`(35.76*g)/(10.81*g*mol^-1)` `xx` `6.022xx10^23` `mol^-1` `=` `??`

Answer 2

`1.991xx10^24`

Explanation:

Molar mass of B = 10.81g/mol
So, 1 mole of B = 10.81g
and 1 mole = `N_A`= `6.023xx10^23`
By combining,
10.81g of B = `6.023xx10^23` atoms of B
thus,
35.76g of B = `(6.023xx10^23)/(10.81)xx35.76`
35.76g of B = `1.991xx10^24`