Question

Calculate the number of cations and anions in each of the following compounds. Enter your answers in scientific notation. (a) 6.42 g of KBr:

Answer 1

Answer: The number of cations are
`3.24 * 10^(22)` and number of anions are
`3.24 * 10^(22)` in 6.42 g of KBr.

Step-by-step explanation:

The molar mass of KBr is (39.10 + 79.90) g/mol = 119.00 g/mol

Now, the dissociation equation for KBr is as follows.

`KBr \rightarrow K^(+) + Br^(-)`

This means that 1 mole of KBr is forming 1 mole of
`K^(+)` (cation) and 1 mole of
`Br^(-)` (anion).

According to mole concept, 1 mole of every substance contains
`6.022 * 10^(23)` atoms. Hence, number of cations present in 6.42 g KBr is calculated as follows.

`No. of cations = Moles * 6.022 * 10^(23)\\= (mass)/(molar mass) * 6.022 * 10^(23)\\= (6.42 g)/(119.00 g/mol) * 6.022 * 10^(23)\\= 3.24 * 10^(22)`

As according to the equation, there are equal number of moles of both cation and anions.

This means that the number of anions are also
`3.24 * 10^(22)`.

Thus, we can conclude that the number of cations are
`3.24 * 10^(22)` and number of anions are
`3.24 * 10^(22)` in 6.42 g of KBr.