Question

How many atoms of oxygen are present in 300 grams of CaCO3? (a) 54.207 × 1023 (b) 6.207 × 1023 (c) 12.207 × 1023 (d) 22.2 × 1023

Answer 1

Answer:

Number of atoms of oxygen present = (A) 54.208 * 10²³ atoms

Explanation:

$\blacksquare$ We know that,

$\implies$ Number of moles = $\dfrac{Given \ \ mass}{Molar \ \ mass}$

$\implies$ Number of moles = $\dfrac{Number \ \ of \ \ molecules}{Avogadro \ \ number}$

$\blacksquare$ Therefore,

$\dfrac{Given \ \ mass}{Molar \ \ mass}= \dfrac{Number\ \ of \ \ molecules}{Avogadro \ \ number}$

$\blacksquare$ Given that,

$\implies$ Given mass = 300 g

$\implies$ Molar mass = 40 + 12 + 48 = 100 g

$\implies$ Avogadro number = 6.022 * 10²³

$\blacksquare$ Substituting these values in we get,

$\dfrac{300}{100} = \dfrac{Number \ \ of \ \ molecules}{6.022 * 10^{23}}$

Number of molecules = 3 * 6.022 * 10²³

Number of molecles = 18.066 * 10²³

1 CaCO₃ molecule has 3 oxygen atoms

18.066 * 10²³ CaCO₃ molecules has  3 * 18.066 * 10²³ atoms

                                                          = 54.208 * 10²³ atoms

Answer 2

Explanation:

Number of atoms of oxygen present = (A) 54.208 * 10²³ atoms

Explanation:

\blacksquare■ We know that,

\implies⟹ Number of moles = \dfrac{Given \ \ mass}{Molar \ \ mass}

Molar mass

Given mass

\implies⟹ Number of moles = \dfrac{Number \ \ of \ \ molecules}{Avogadro \ \ number}

Avogadro number

Number of molecules

\blacksquare■ Therefore,

\dfrac{Given \ \ mass}{Molar \ \ mass}= \dfrac{Number\ \ of \ \ molecules}{Avogadro \ \ number}

Molar mass

Given mass

=

Avogadro number

Number of molecules

\blacksquare■ Given that,

\implies⟹ Given mass = 300 g

\implies⟹ Molar mass = 40 + 12 + 48 = 100 g

\implies⟹ Avogadro number = 6.022 * 10²³

\blacksquare■ Substituting these values in we get,

\dfrac{300}{100} = \dfrac{Number \ \ of \ \ molecules}{6.022 * 10^{23}}

100

300

=

6.022∗10

23

Number of molecules

Number of molecules = 3 * 6.022 * 10²³

Number of molecles = 18.066 * 10²³

1 CaCO₃ molecule has 3 oxygen atoms

18.066 * 10²³ CaCO₃ molecules has 3 * 18.066 * 10²³ atoms

= 54.208 * 10²³ atoms