# Question 560e6

Dec 7, 2017

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

#### Explanation:

The quotient $\frac{89.1 \cdot g}{208.23 \cdot g \cdot m o {l}^{-} 1} = 0.428 \cdot m o l$ gives us the molar quantity....and we know that there are moles of atoms per mole of barium chloride....$2 \cdot m o l \times C {l}^{-}$ ion, and $1 \cdot m o l \cdot B {a}^{2 +}$ ion.

And so we take the product...$6.022 \times {10}^{23} \cdot m o {l}^{-} 1 \times 0.428 \cdot m o l \times 3 = 7.73 \times {10}^{23} \cdot \text{atoms}$

Dec 7, 2017

$2.577 \cdot {10}^{23}$ molecules of ${\text{BaCl}}_{2}$ - or $7.73 \cdot {10}^{23}$ atoms.

#### Explanation:

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

This tells me that 1 mole of ${\text{BaCl}}_{2}$ is 208.23 grams per mole.

89.1 g is 0.428 of one mole (rounding)

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

$0.428 \cdot 6.023 \cdot {10}^{23} = 2.577 \cdot {10}^{23}$ molecules of ${\text{BaCl}}_{2}$.

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

GOOD LUCK