Question

1. How many molecules of barium hydroxide, Ba(OH)2 are there in 21.9 grams of

barium hydroxide?
A. 6.022 x 1023 molecules Ba(OH)2
B. 7.7 x 1022 molecules Ba(OH)2
C. 6.23 x 1021 molecules Ba(OH)2
D. 5.93 x 1021 molecules Ba(OH)2

Answer 1

Answer:
`7.7* 10^(22)` molecules of
`Ba(OH)_2`

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}}=(21.9g)/(171g/mol)=0.128moles`

1 mole of
`Ba(OH)_2` contains =
`6.023* 10^(23)` molecules

0.128 mole of
`Ba(OH)_2` contains =
`(6.023* 10^(23))/(1)* 0.128=7.71* 10^(22)` molecules

Thus there are
`7.7* 10^(22)` molecules of
`Ba(OH)_2`