Answer
a.) $4.785\times10^{23}$ Formula units $K_{2}CrO_{4}$
b.) $9.570\times10^{23}$ $K^{+}$ ions
c.) $4.785\times10^{23}$ $CrO^{2-}_{4}$ ions
d.) $3.350\times10^{24}$ atoms
Work Step by Step
a.) We can find the number of formula units present in 154.3 grams of $K_{2}CrO_{4}$ by converting the given mass into moles, then multiplying the product by the Avogardo's number ($6.022\times10^{23}$) as shown below:
Formula units $K_{2}CrO_{4}$ = $154.3 g K_{2}CrO_{4} \times \frac{1 mol K_{2}CrO_{4}}{194.20 g K_{2}CrO_{4}} \times \frac{6.022\times10^{23} Formula units K_{2}CrO_{4}}{1 mol K_{2}CrO_{4}} = 4.785\times10^{23} Formula units K_{2}CrO_{4}$
b.) We know that there are two $K^{+}$ ions in one formula unit $K_{2}CrO_{4}$. Therefore,
$K^{+} ions = 4.785\times10^{23} Formula units K_{2}CrO_{4} \times \frac{2 K^{+} ions}{1 Formula unit K_{2}CrO_{4}} = 9.570\times10^{23} K^{+} ions$
c.) We also know that there's one $CrO^{2-}_{4}$ ion for every formula unit $K_{2}CrO_{4}$. Therefore,
$CrO^{2-}_{4} ions = 4.785\times10^{23} Formula units K_{2}CrO_{4} \times \frac{1 CrO^{2-}_{4} ion}{1 Formula unit K_{2}CrO_{4}} = 4.785\times10^{23} CrO^{2-}_{4} ions$
d.) In a single molecule of $K_{2}CrO_{4}$, there's a total of 7 atoms present: 2 potassium atoms, 1 Cr atom, and 4 oxygen atom. Similarly, there are also 7 atoms in one formula unit $K_{2}CrO_{4}$.
Number of atoms = $4.785\times10^{23} Formula units K_{2}CrO_{4} \times \frac{7 atoms}{1 Formula unit K_{2}CrO_{4}} = 3.350\times10^{24} atoms$
