Answer to Question #91804 in General Chemistry for jimbo

Question

How many atoms are in
37.9
37.9
g of potassium bromite (KBrO2)

Answer 1

find the molar mass of the substance: M(KBrO₂)=M(K)+M(Br)+2M(O)=151g/mol

Now find the number of mollecules of the substance: N_m=(m(KBrO₂₎/M(KBrO₂))*N_A≈1.5*10²³ (N_A -Avogadro constant )

In this mollecule, the number of atoms is 4 times greater than the number of mollecules: N=N_m*4=6*10²³;

Answer: 6*10²³