Question

Determine the numbers of atoms in each of the following 5.40 g B 0.250 mol k 0.0384 mol k 0.02550 g pt 1.00 x 10^-10 g Au,

Answer 1

a) 5.40 g B :
`3.012* 10^(23)` atoms

b) 0.250 mol K :
`1.51* 10^(23)` atoms

c) 0.0384 mol K :
`0.23* 10^(23)` atoms

d) 0.02550 g Pt:
`7.8* 10^(19)` atoms

e)
`1.00* 10^-{10} g` Au:
`0.03* 10^(13)` atoms

Step-by-step explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number
`6.023* 10^(23)` of particles.

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}`

a) 5.40 g B

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=(5.40g)/(11g/mol)=0.5moles`

1 mole of boron contains =
`6.023* 10^(23)` atoms

Thus 0.5 moles of boron contain =
`(6.023* 10^(23))/(1)* 0.5=3.012* 10^(23)` atoms

b) 0.250 mol K

1 mole of potassium (K) contains =
`6.023* 10^(23)` atoms

Thus 0.250 moles of potassium contain =
`(6.023* 10^(23))/(1)* 0.250=1.51* 10^(23)` atoms

c) 0.0384 mol K

1 mole of potassium (K) contains =
`6.023* 10^(23)` atoms

Thus 0.0384 moles of potassium (K) contain =
`(6.023* 10^(23))/(1)* 0.0384=0.23* 10^(23)` atoms

d) 0.02550 g Pt

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=(0.02550 g)/(195g/mol)=1.3* 10^(-4)moles`

1 mole of platinum contains =
`6.023* 10^(23)` atoms

Thus
`1.3* 10^(-4)moles` of platinum contain=
`(6.023* 10^(23))/(1)* 1.3* 10^(-4)=7.8* 10^(19)` atoms

e)
`1.00* 10^-{10}g` Au,

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=(1.00* 10^(-10)g)/(197g/mol)=0.005* 10^(-10)moles`

1 mole of gold contains =
`6.023* 10^(23)` atoms

Thus
`0.005* 10^(-10)moles` of platinum contain=
`(6.023* 10^(23))/(1)* 0.005* 10^(-10)=0.03* 10^(13)` atoms

Answer 2

The answers are the following:
1. 3.01 x 10^23 atoms B
solution: (5.40/10.81)(6.022x10^23)
2. 1.51 x 10^23 atoms S
solution: (.250)(6.022x10^23)
3. 2.31 x 10^22 atoms K
solution: (.0384)(6.022x10^23)
4. 7.872 x 10^19 atoms Pt
solution: (.02550/195.08)(6.022x10^23)
5. 3.06 x 10^11 atoms Au
solution: (1.00x10^-10/196.97)(6.022x10^23)