Question

How many cations are present in 28.6 g of silver chromate?

How many anions are present?

What is the mass in grams of one formula unit of silver chromate?

Answer 1

Answer:

Number of silver ions : $1.037\times 10^{23}$

Number of chromate ions :$5.18 \times 10^{22}$

Mass of single silver chromate : $5.51\times 10^{-22} g$

Explanation:

$Ag_2CrO_4$ , Silver chromate

Mass of silver chromate = 28.6 g

Molar mass of silver chromate = 332 g/mol

Moles of silver chromate = $\frac{28.6 g}{332 g/mol}=0.0861 mol$

$Ag_CrO_4\rightarrow 2Ag^++CrO_4^{2-}$

1 mole of silver chromate has 2 mole of silver ions and 1 mole of chromate ions.

Then silver ions in 0.0861 moles of silver chromate :

$2\times 0.0861 moles=0.1722 moles$ of silver ion

Then chromate ions in 0.0861 moles of silver chromate :

$1\times 0.0861 moles=0.0861 moles$ of chromate ion

Number of silver ions :

$=0.1722\times 6.022\times 10^{23}=1.037\times 10^{23}$

Number of chromate ions :

$=0.0861\times 6.022\times 10^{23}=5.18 \times 10^{22}$

Mass of 1 mole of silver chromate = 332 g

1 mole= $6.022\times 10^{23}$ molecules/atoms/ions

Mass of single silver chromate :

$\frac{332 g}{6.022\times 10^{23}}=5.51\times 10^{-22} g$