Question

Determine the number of milligrams of lead (II) nitrate that would contain 6.894 *10¹⁹ atoms of oxygen. (give you answer with 3 decimal places and no units)

Answer 1

Pb(NO3)2

6 atoms of oxygen contains by 1 molecule

1 atoms of oxygen contains = 1/6 molecule of Pb(NO3)2

6.894*10^19 atoms will contains = 1/6 * ( 6.894*10^19)

= 1.149*10^19

Moles of Pb(NO3)2 = total molecule / avagadro number

= (1.149*10^19) / ( 6.022*10^23)

= 1.91*10^-5

Mass of Pb(NO3)2 = moles * molar mass

= 1.91*10^-5 * 331.2

= 6.319*10^-3 grams

Or 6.319 mg

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