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

2 Answers
Nov 15, 2015

Answer:

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# #=# #??#

Nov 15, 2015

Answer:

#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#