How many sodium atoms are there in 6.0 g of Na3N?
I am doing this problem by two methods:
1)
Molecular weight for Na3N = 83 g/mol
6.0g /( 83g/mol) = 0.0723 mol Na3N
0.0723 mol Na3N * 3mol Na/1mol Na3N = 0.217 mol Na
one mole contain= 6.022*1023 atoms (avagadro
number)
0.217 mol * 6.022*10^23 atoms/mol = 1.306*10^23 sodium atoms in 6
grams of Na3N
2)
%Na in Na3N = (23x3)gmol^-1 / [(23X3) + 14]gmol^-1 =
83.1325301%
in 6g --> mass of nitrogen = (83.1325301/100) x 6g =
4.987951807g
n=m/M
n=moles
m=mass(g)
M=molar mass(gmol^-1)
n(N) = 4.987951807g / 23gmol^-1 = 0.21686747 moles
1 mole of nitrogen atoms = 6.022x10^23 nitrogen atoms
0.21686747 moles = 0.21686747 moles x 6.022x10^23
= 1.30598 x 10^23 nitrogen atoms
thus, answer = c. 1.3 × 10^23 atoms
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