Question

how many atoms of phosphorous are in 4.90 mol of copper(ii) phosphate? the formula for copper(ii) phosphate is Cu3(PO4)2.

Answer 1

Answer:

$\boxed{5.90 \times 10^{24}}$

Explanation:

Step 1. Calculate the formula units of Cu₃(PO₄)₂  

$\text{4.90 mol Cu _{3} (PO _{4} ) _{2} } \times \dfrac{6.022 \times 10^{23} \text{ formula units}}{\text{1 mol Cu _{3} (PO _{4} ) _{2} }}\\\\=\text{2.951 \times 10 ^{24} formula units Cu _{3} (PO _{4} ) _{2} }$

Step 2. Calculate the atoms of P

$\text{Atoms of P}\\\\=\text{2.951 \times 10 ^{24} formula units Cu _{3} (PO _{4} ) _{2} } \times \dfrac{ \text{2 atoms P}}{\text{1 formula unit Cu _{3} (PO _{4} ) _{2} }}\\\\= \boxed{5.90 \times 10^{24} \text{ atoms P}}$