A subtance X contains 10 gram of calcium carbonate calculate the number of oxygen atom present in 10 gram of calcium carbonate ​

[tex] \LARGE{ \boxed{ \rm{ \purple{Required \: answer}}}}[/tex]

☃️ Chemical formulae ➝ [tex]\sf{CaCO_3}[/tex]

How to find?

For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.

[tex] \boxed{ \sf{No. \: of \: moles = \frac{given \: weight}{molecular \: weight} }}[/tex]

And, Then we need to find the number of molecules which can be found by:

[tex] \large{ \boxed{ \sf{no. \: of \: molecules = N_A \times no. \: of \: moles}}}[/tex]

Here, Na = Avagadro Number (6.022 × 10²³)

☀️ As it is a compund, We considered molecules. Then, we can predict the no. of atoms of elements in one molecule of compound and get the answer.

Solution:

Atomic weight of elements:

❍ Molecular weight of [tex]\sf{CaCO_3}[/tex]

= 40 + 12 + 3 × 16

= 52 + 48

= 100 g/mol

❍ Given weight: 10 g

Then, no. of moles,

⇛ No. of moles = 10 g / 100 g mol‐¹

⇛ No. of moles = 0.1 moles

☄ No. of moles of Calcium carbonate in the given weight = 0.1 moles

Finding no. of molecules,

⇛ No. of molecules of CaCO3 = 0.1 × 6.022 × 10²³

⇛ No. of molecules of CaCO3 = 6.022 × 10²²

Now,

Then, 6.022 × 10²² molecules will have,

= 3 × 6.022 × 10²² atoms of oxygen

= 18.022 × 10²² atoms of oxygen

In more scientific form,

= 1.8022 × 10²³ atoms of oxygen

☄ Hence, Final answer - 1.8022 × 10²³ atoms

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