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 \(CO_2\):

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 \(CO_2\):

\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