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

The mass of sulfur that would have precisely 4.7 × 10²² atoms of sulfur is

2.497 grams.

The Avogadro law states that an equal volume of all gases at the same temperature and pressure contains the same number of molecules.

Recall that:

the molar mass of sulfur = 32 g/mol

It implies that 1 mole of sulfur is composed of 32 g weight which contains 6.023 × 10²³ atoms

If 1 mole of sulfur = 32 g = 6.023 × 10²³ atoms

∴

The mass of sulfur that is contained in 4.7 × 10²² atoms of sulfur is estimated as follows:

x(g) = 2.497 grams

Therefore, we can conclude that the mass of sulfur that would have precisely 4.7 × 10²² atoms of sulfur is 2.497 grams.

Learn more about the sulfur here:

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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

thats how its supposed to be

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b