# Question c2d91

Feb 17, 2018

Around $1.06 \cdot {10}^{24}$ ions

#### Explanation:

$O {H}^{-}$ has a molar mass of $17 \setminus \text{g/mol}$.

So, in $30 \setminus \text{g}$ of hydroxide ions. there exists (30cancel"g")/(17cancel"g""/mol")~~1.76 \ "mol"#

In one mole of molecules, there are $6.02 \cdot {10}^{23}$ molecules.

So, there will be a total of

$1.76 \cdot 6.02 \cdot {10}^{23} \approx 1.06 \cdot {10}^{24}$ hydroxide ions.

Feb 17, 2018

See below

#### Explanation:

We know that in 1 Mole we have $6.022 \cdot {10}^{23}$ particles.

In this situation, we aren't given the specified amount of moles, so we have to work them out using the formula:

moles $= \frac{m a s s}{M} _ r$

Where the ${M}_{r}$ is the relative formula mass.

Here we have a mass of $30$ and an ${M}_{r}$ of $17$

$\therefore$ the amount of moles we have is $\frac{30}{17}$ moles.

To work out the number of particles, we multiply the number of moles by Avogadro's constant $\left(6.022 \cdot {10}^{23}\right)$

$\therefore \frac{30}{17} \cdot 6.022 \cdot {10}^{23} = 1.062705882 \cdot {10}^{23}$ ions