Calculation based on number of atoms — lesson. Science State Board, Class 10.

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1.	Calculate the number of atoms present in $1$ gram of gold (atomic mass of $Au=198$ ).

Data:

Given mass of gold

$=1$

$g$

Atomic mass of

$Au=198$

Avogadro's number

$=6.023\times10^{23}$

Solution:

$\begin{array}{l} Number of atoms of Au & = & \frac{Mass of Au \times Avogadro number}{Atomic mass of Au} \\ = & \frac{1}{198} \times 6.023 \times 10^{23} \\ = & 3.042 \times 10^{21} g \end{array}$

Therefore, the number of atoms present in

$1$ gram of gold

$=3.042\times10^{21}$

$g$

2.	Calculate the number of atoms of oxygen and carbon in $5$ $moles$ of $CO_2$ .

Calculating number of oxygen atoms in

$5$

$moles$ of

$CO_2$ :

$1$ $mole$ of $CO_2$ contains $2$ $moles$ of oxygen.

$5$ $moles$ of $CO_2$ contain $10$ $moles$ of oxygen.

Solution:

$\begin{array}{l} No . of atoms of O_{2} & = & No . of moles of O_{2} \times Avogadro number \\ = & 10 \times {6.023 \times 10}^{23} \\ = & {6.023 \times 10}^{24} \end{array}$

Therefore, the number of atoms of oxygen in

$5$

$moles$ of

$CO_2=6.023\times10^{24}$

Calculating the number of carbon atoms in

$5$

$moles$ of

$CO_2$ :

$1$ $mole$ of $CO_2$ contains $1$ $moles$ of carbon.

$5$ $moles$ of $CO_2$ contain $5$ $moles$ of carbon.

$\begin{array}{l} No . of atoms of C & = & No . of moles of C \times Avogadro number \\ = & 5 \times {6.023 \times 10}^{23} \\ = & {3.011 \times 10}^{24} \end{array}$

Therefore, the number of atoms of carbon in

$5$

$moles$ of

$CO_2=3.011\times10^{24}$

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9. Calculation based on number of atoms

Theory: