Question #560e6

2 Answers
anor277
Dec 7, 2017

Well, 3xxN_Axx(89.1*g)/(208.23*g*mol^-1)=??

Explanation:

The quotient (89.1*g)/(208.23*g*mol^-1)=0.428*mol gives us the molar quantity....and we know that there are moles of atoms per mole of barium chloride....2*molxxCl^- ion, and 1*mol*Ba^(2+) ion.

And so we take the product...6.022xx10^23*mol^-1xx0.428*molxx3=7.73xx10^23*"atoms"

jim p. ยท Stefan V.
Dec 7, 2017

2.577 * 10^23 molecules of "BaCl"_2 - or 7.73 * 10^23 atoms.

Explanation:

I had to use Wolfram to find the atomic weight of Barium Chloride.

Wolfram Alpha

This tells me that 1 mole of "BaCl"_2 is 208.23 grams per mole.

89.1 g is 0.428 of one mole (rounding)

1 mole is 6.023 * 10^23 molecules, which is Avogadro's number. Which is one of the 2 numbers I can remember from my schooling so many years ago.

0.428 * 6.023 * 10^23 = 2.577 * 10^23 molecules of "BaCl"_2.

Each molecule is 1 Barium and 2 chlorine, a total of 3 atoms, so multiply the number of molecules by 3 gives you 7.73 * 10^23 atoms.

GOOD LUCK

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