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) Mass of
`3.0* 10^(23) atoms` of mercury is 99.91 g.

b)
`3.0021* 10^(12) atoms` in 1 nano gram of mercury.

Step-by-step explanation:

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

Number of molecules or number of atoms:

`n* N_A`

`N_A=6.022* 10^(23)` atoms or molecules

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

Number of mercury atoms =
`3.0* 10^(23) atoms`

Moles of Mercury :

`(3.0* 10^(23) atoms)/(6.022* 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* 10^(23) atoms` of mercury is 99.91 g.

b) Mass of mercury = 1 nano gram =
`10^(-9) g`

Moles of mercury:
`(10^(-9) g)/(200.59 g/mol)`

Number of atoms of mercury in 1 nano gram:

:
`(10^(-9) g)/(200.59 g/mol)* 6.022* 10^(23)`

`=3.0021* 10^(12) atoms`

Answer 2

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