Question

Find the ratio of atoms present in 16 g of o2 and 32 g of o3

Answer 1

Answer:

Explanation:

1:3

Answer 2

Answer:  

$1:2$

Explanation:

Given: Here we have $16$ g of $O_{2}$ and $32$ g of $O_{3}$.

To find: We have to find the ratio of atoms present in $16$ g of $O_{2}$ and $32$ g of $O_{3}$.

Concept:  The mole idea is a useful way to indicate how much of a substance there is. Any measurement can be divided into two components: The magnitude in number and the units in which the magnitude is expressed.

Step 1

Mass of  $O_{2}$ is $=$  $16$ g

Mass of $O_{3}$ is $=$ $32$ g

The molar mass of $O_{2}$ is  $=$ $32$ g

The molar mass of $O_{3}$ is $=$ $48$ g

Step 2

Then, the number of moles of $O2$ = $\frac{16}{32}$

                                                       = $\frac{1}{2}$ mole of $O_{2}$

Now, the number of moles of $O_{3}$ = $\frac{32}{48}$

                                                      = $\frac{2}{3}$ mole of $O_{3}$

Step 3

As we know that 1 mole of substances contains $6.022$×$10^{23}$ atoms.

So, the $\frac{1}{2}$ mole of $O_{2}$ will contains = $\frac{1}{2}$×$6.022$×$10^{23}$×$2$

And the $\frac{2}{3}$ moles of $O_{3}$ will contains = $\frac{2}{3}$×$6.022$×$10^{23}$×$3$

Now, the ratio of the number of atoms present in it  = $\frac{1}{2}$

Hence, the ratio will be $1:2$.

≠SPJ2