What mass of sulfur would have precisely 4.7x10^22 atoms of sulfur?

2.498g

Explanation:

First off it's important to know Avogadro's number relationship with number of moles and molar mass.

It is given as;

1 mole = 6.02 * 10^23 atoms.

A mole's weight is equal to it's molar mass. Hence the equation can be changed to;

1 mole = molar mass = 6.02 * 10^23 atoms

In this problem, molar mass of Sulphur is 32g/mol. This means 1 mole of sulphur weighs 32 g and contains 6.02 * 10^23 atoms. What mass of sulphur would then contain 4.7x10^22 of atoms?

This leads us to;

32 g = 6.02 * 10^23

x g = 4.7x10^22

Upon cross multiplication, we have;

x = (4.7x10^22 * 32) / 6.02 * 10^23

x = 24.98 * 10^-1

x = 2.498g

Answer: It would be 2.5