Stephenvion
Stephenvion Stephenvion
  • 02-09-2016
  • Chemistry
contestada

What is the atomic weight of an element if 4 grams of it contains 2.98 x 10^22 atoms?

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MysteriousGamer
MysteriousGamer MysteriousGamer
  • 08-09-2016
n=amount  in moles
N=amount in numbers=2.98x10^22
L=Avagadro's constant=mole-1(6.022x10^23)
m= mass in grams
M=molar mass in g/mol
Now,
n=N/L
  =2.98x10^22/6.022x10^23
  =0.0495 mol ≈0.05 mol
Then,
M=m/n
   =4/0.05
   =80 g/mol
The element is Br
Answer Link
HomertheGenius
HomertheGenius HomertheGenius
  • 12-09-2016
m = 4 g;  N = 2.98 * 10^22
Avogadro constant:  L = 6.022 * 10^23
Number of moles:
n = 2.98 * 10^22  : 6.022 * 10^23 = 0.298 * 10^23 : 6.022 * 10^23 =
= 0.0495 moles
Molar mass:
M = m/n = 4 g / 0.0495 = 80.8 ≈ 80  
Answer Link

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