Respuesta :
a.200.59 amu of mercury contains = 6.02 * 1023 molecules
Mass for 3.0 * 1023 molecules = (3.0 * 1023 * 200.59) /6.02 * 1023
= 100.29 amu
Thus, mass for 3.0 * 1023 molecules is 100.29 amu
b.As, 200.59 g mercury contains = 6.02 * 1023 atoms
1 g mercury contains = (6.02 * 1023 /200) atoms
= 3.01 * 1021 atoms
Thus, 1 ng mercury contains = (3.01 * 1021) / 109
= 3.01 * 1012 atoms
Thus, 3.0 * 1012 atoms present in 1 ng mercury
Answer:
a) Mass of [tex]3.0\times 10^{23} atoms[/tex] of mercury is 99.91 g.
b)[tex]3.0021\times 10^{12} atoms[/tex] in 1 nano gram of mercury.
Explanation:
[tex]Moles(n)=\frac{\text{mass of compound}}{\text{Molar mass of compound}}[/tex]
Number of molecules or number of atoms:
[tex]n\times N_A[/tex]
[tex]N_A=6.022\times 10^{23} [/tex] atoms or molecules
Mercury has an atomic mass of 200.59 amu = 200.59 g/mol
Number of mercury atoms = [tex]3.0\times 10^{23} atoms[/tex]
Moles of Mercury :
[tex]\frac{3.0\times 10^{23} atoms}{6.022\times 10^{23}}=0.4981 moles[/tex]
Mass of 0.4981 moles of mercury:
0.4981 moles × 200.59 amu = 99.91 g
Mass of [tex]3.0\times 10^{23} atoms[/tex] of mercury is 99.91 g.
b) Mass of mercury = 1 nano gram = [tex]10^{-9} g[/tex]
Moles of mercury:[tex]\frac{10^{-9} g}{200.59 g/mol}[/tex]
Number of atoms of mercury in 1 nano gram:
:[tex]\frac{10^{-9} g}{200.59 g/mol}\times 6.022\times 10^{23}[/tex]
[tex]=3.0021\times 10^{12} atoms[/tex]
