Avogrado’s Number: The Mole - if you could answer one or all & show me how...?

Question

Avogrado’s Number: The Mole
2.45) Use a factor derived from the relationships written in Exercise 2.43 and the factor-unit method to determine the following:

a) the number of moles of potassium atoms in a 50.0-g sample of potassium

b) the number of magnesium atoms in a 1.82-mol sample of magnesum

c) the number of tin atoms in a 200-g sample of argon

Answer 1

I see some answers, but let's talk about how to do it:

What is a mole? A mole is the truly splendiferous number (6.022 x 10^23) of atoms it takes of an element to equal the that element's atomic weight on the periodic table.

The atomic weight of Potassium is 39.0983 g.

So Potassium is 39.0983 grams per 1 mole. And you have 50 grams - so from the start, you know that you should be dealing with more than 1 mole of Potassium. 50.0 grams / 39.0983 grams/mole= 1.3 moles Potassium

Answer 2

a) Potassium has an atomic weight of 39.0983 grams/mole. You have a 50.0 gram sample of pure (I assume) potassium metal. Therefore you have:

(50.0 g)/(39.0983 g/mole) = 1.279 moles of Potassium

this would also be equal to:

(1.279 moles) * (6.023E23 atoms/mole) = 7.703E23 atoms

that's ~770,300,000,000,000,000,000,000 atoms of potassium

Question B can be answered by the same process.

Question C is a trick question -- there should be zero atoms of tin in 200 grams of argon gas.

Answer 3

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Answer 4

What did you come up with for Exercise 2.43? Use that...

This should be easy division/multiplication after you look up the info on K & Mg. It looks like 2.45c is goofed up, as Sn != Ar.