Respuesta :
The atomic mass of each element from the periodic table can help you find those masses. The atomic mass is the molar mass of 6.02 x 10^23 atoms of that element, so...
A. Atomic mass of Au is 196.97 g/mol. Dividing by the Avogadro constant (6.02 x 10^23), or 1 mol of atoms, would give you the weight in grams for one atom of gold: 3.27 x 10^-22
B. The atomic mass of mercury is 200.59 g/mol, so one mol of mercury would weigh 200.59 g.
C. Chlorine weighs 35.45 g/mol, so two mols would be twice that, or 70.90 g.
D. We know that 6.02 x 10^23 atoms are more than 25 atoms of mercury, so D would already be far less than B, but let's do the calculations just in case. 1 mol of Hg weighs 200.59 g/mol, divide by Avogadro's number to get the weight per atom (3.33 x 10^-22 g), and multiply by 25 - 8.33 x 10^-21 g.
By the results, we can see that 1 mol of Hg has the greatest mass of all the possible choices. Generally, the weight of one or a few single atoms is minuscule, so we could have eliminated A and D immediately, but it's good to understand and do the math,
A. Atomic mass of Au is 196.97 g/mol. Dividing by the Avogadro constant (6.02 x 10^23), or 1 mol of atoms, would give you the weight in grams for one atom of gold: 3.27 x 10^-22
B. The atomic mass of mercury is 200.59 g/mol, so one mol of mercury would weigh 200.59 g.
C. Chlorine weighs 35.45 g/mol, so two mols would be twice that, or 70.90 g.
D. We know that 6.02 x 10^23 atoms are more than 25 atoms of mercury, so D would already be far less than B, but let's do the calculations just in case. 1 mol of Hg weighs 200.59 g/mol, divide by Avogadro's number to get the weight per atom (3.33 x 10^-22 g), and multiply by 25 - 8.33 x 10^-21 g.
By the results, we can see that 1 mol of Hg has the greatest mass of all the possible choices. Generally, the weight of one or a few single atoms is minuscule, so we could have eliminated A and D immediately, but it's good to understand and do the math,
