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

Question

Patience Garza

Chemistry

3 January, 21:44

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

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Answers (2)

Know the Answer?

Robbins · Answer 1

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

Duran · Answer 2

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