Question

10 mole of Ba(NO3)2 dissolves in water to give how many NO3- ions?
1 - 6.02 x 10 to the 23
2 - 6.02 x 10 to the minus 23
3 - 6.02 x 10 to the 24
4 - 1.204 x 10 to the 23
5 - 1.204 x 10 to the 25

What I want to know is how do I work out the question.

Answer 1

Answer on Question #43040 - Chemistry - Other

Question:

10 mole of Ba(NO3)2 dissolves in water to give how many NO3- ions?

1 - 6.02 x 10 to the 23

2 - 6.02 x 10 to the minus 23

3 - 6.02 x 10 to the 24

4 - 1.204 x 10 to the 23

5 - 1.204 x 10 to the 25

What I want to know is how do I work out the question.

Answer:

Dissolution equation for $\mathrm{Ba(NO_3)_2}$:

$$\mathrm{Ba(NO_3)_2} = \mathrm{Ba^{2+}} + 2\mathrm{NO_3^-}$$

We see that each molecule of barium nitrate dissociates producing $2\mathrm{NO_3^-}$ ions. So, one mole of $\mathrm{Ba(NO_3)_2}$ produces 2 moles of $\mathrm{NO_3^-}$ ions, and 10 mol of $\mathrm{Ba(NO_3)_2}$ produce 20 mol of $\mathrm{NO_3^-}$ ions.

One mole of any substance has $6.02 \times 10^{23}$ particles (atoms, molecules or ions; Avogadro constant). So, we make a proportion:

1 mol has $6.02 \times 10^{23}$ particles

20 mol of $\mathrm{NO_3^-}$ ions - $x\mathrm{NO_3^-}$ ions

$$x = \frac{20 \cdot 6.02 \cdot 10^{23}}{1} = 1.204 \cdot 10^{25}$$

So, 10 mole of $\mathrm{Ba(NO_3)_2}$ dissolves in water giving $1.204 \cdot 10^{25} \mathrm{NO_3^-}$ ions.

Answer: 5 - 1.204 x 10 to the 25.

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