Question #9908

How many potassium atoms are in 6.5g of K3PO4?

Expert's answer

Molar mass of K3PO4 is M(k3PO4)=3*39+31+4*16=212 g/mol. The quantity of K3PO4 is

n(K3PO4)=6.5g/212g/mol=0.0307mol. The quantity of K is n(K)=3*n(K3PO4)=0.092mol

or N(K)=n(K)*(Avogadro's number)=0.092*6.022*10^23=5.54*10^22 atoms.

n(K3PO4)=6.5g/212g/mol=0.0307mol. The quantity of K is n(K)=3*n(K3PO4)=0.092mol

or N(K)=n(K)*(Avogadro's number)=0.092*6.022*10^23=5.54*10^22 atoms.

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