Question

32. Mercury has an atomic mass of 200.59 amu. Calculate the a. Mass of 3.0 x 1023 atoms. b. Number of atoms in one nanogram of Mercury? i needed help on this question please help

Answer 1

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 2

Answer:

a) Mass of $3.0\times 10^{23} atoms$ of mercury is 99.91 g.

b)$3.0021\times 10^{12} atoms$ in 1 nano gram of mercury.

Explanation:

$Moles(n)=\frac{\text{mass of compound}}{\text{Molar mass of compound}}$

Number of molecules or number of atoms:

$n\times N_A$

$N_A=6.022\times 10^{23}$ atoms or molecules

Mercury has an atomic mass of 200.59 amu = 200.59 g/mol

Number of mercury atoms = $3.0\times 10^{23} atoms$

Moles of Mercury :

$\frac{3.0\times 10^{23} atoms}{6.022\times 10^{23}}=0.4981 moles$

Mass of 0.4981 moles of mercury:

0.4981 moles × 200.59 amu = 99.91 g

Mass of $3.0\times 10^{23} atoms$ of mercury is 99.91 g.

b) Mass of mercury = 1 nano gram = $10^{-9} g$

Moles of mercury:$\frac{10^{-9} g}{200.59 g/mol}$

Number of atoms of mercury in 1 nano gram:

:$\frac{10^{-9} g}{200.59 g/mol}\times 6.022\times 10^{23}$

$=3.0021\times 10^{12} atoms$