Question

What is the atomic mass of an element if 4.00 grams of it contains 2.98x1022 atoms ?

Answer 1

Answer:

Atomic mass of the element = 80.80 amu

Explanation:

Given:

Mass of element = 4.00 g

No. of atoms = $2.98\times 10^{22}$

$mol = \frac{No.\ of\ atoms}{Avogadro's\ no.}$

Avogadro's no. = tex]6.02\times 10^{23}[/tex]

$mol = \frac{No.\ of\ atoms}{Avogadro's\ no.}$

=$\frac{2.98\times 10^{22}}{6.02\times 10^{23}} = 0.0495\ mol$

Formula for moles in the terms of mass and atomic mass

$mol = \frac{Mass\ in\ g}{Atomic\ mass} \\Atomic\ mass = \frac{Mass\ in\ g}{mol}$

Atomic mass of the element = $\frac{4.00}{0.0495} = 80.80\ amu$

Atomic mass of the element = 80.80 amu

Answer 2

Hope this helps you.