Question

The number of moles in $19.86$ g of $Pb(N{O}_3{)}_2$ (Molecular Weight $=207.2$ g/mole) equals

• A. $0.06$ moles
• B. $6.023$ moles
• C. $0.6$ moles
• D. $0.12$ moles

Answer 1

The correct option is A $0.06$ moles

As we know, one mole of each substance contains $N_A=6.023\times {10}^{23}$ molecules/atoms.

Here, moles of $Pb(N{O}_3{)}_2=\frac{19.86}{331.2}=0.06$ moles

Therefore, the number of molecules of $Pb(N{O}_3{)}_2$$=0.06\times 6.023\times {10}^{23}$