Answer
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Hint :For any given ideal gas, one mole of ideal gas occupies $ 22.4litres $ of volume at STP. STP is the standard reference point of temperature $ (273.15k{\text{ or }}{{\text{0}}^ \circ }C) $ and pressure $ (1atm{\text{ or 1}}{{\text{0}}^5}pascals) $ used when measuring gases.
$ 1{\text{ }}mole = 6.022 \times {10^{23}}molecules $
$ mole = \dfrac{{volume}}{{22.4L}} $
Complete Step By Step Answer:
When we are given volume, we can convert it into a number of moles present. This helps us to calculate the number of moles.
$ 1l = 1000cc $
To find mole, the formula is $ mole = \dfrac{{volume}}{{22.4L}} $
Mole $ 1l = 1000cc $
$ mole = \dfrac{{1.12 \times {{10}^{ - 7}} \times {{10}^3}}}{{22.4L}} $
So the number of moles will be:
$ \begin{gathered}
mole = 0.05 \times {10^{ - 4}} \\
= 0.05 \times {10^{ - 4}} \times 6.022 \times {10^{23}}molecules \\
\end{gathered} $ $ $
$ = 3.01 \times {10^{20}} $ molecules.
The molecules are $ = 3.01 \times {10^{20}} $ molecules.
Additional Information:
The two important concepts of science are the mole and Avogadro ’s number that provides a link between the different properties of separate atoms. The number of atoms or other particles in a mole is the same for all the substances. Moles in general is related to the elements mass. We can say that one mole of carbon $ - 12 $ atoms have $ 6.02214076 \times {10^{23}} $ atoms and a mass of 12gram. In comparison to one mole of chlorine, by definition, of the same number of atoms as carbon $ - 12 $ ,but it has a mass of $ 35.453 $ u. When we are given volume, we can convert it into a number of moles present. This helps us to calculate the number of moles. one mole of ideal gas occupies $ 22.4litres $ of volume at STP. STP is the standard reference point of temperature $ (273.15k{\text{ or }}{{\text{0}}^ \circ }C) $ and pressure $ (1atm{\text{ or 1}}{{\text{0}}^5}pascals) $ used when measuring gas.
Note :
The answer will always be in SI unit or in the form the question is asking for. So it is important to convert the units properly according to the question. Focus on whether a question asks for STP or something else. Sometimes the answer is wrong due to incorrect methods applied.
$ 1{\text{ }}mole = 6.022 \times {10^{23}}molecules $
$ mole = \dfrac{{volume}}{{22.4L}} $
Complete Step By Step Answer:
When we are given volume, we can convert it into a number of moles present. This helps us to calculate the number of moles.
$ 1l = 1000cc $
To find mole, the formula is $ mole = \dfrac{{volume}}{{22.4L}} $
Mole $ 1l = 1000cc $
$ mole = \dfrac{{1.12 \times {{10}^{ - 7}} \times {{10}^3}}}{{22.4L}} $
So the number of moles will be:
$ \begin{gathered}
mole = 0.05 \times {10^{ - 4}} \\
= 0.05 \times {10^{ - 4}} \times 6.022 \times {10^{23}}molecules \\
\end{gathered} $ $ $
$ = 3.01 \times {10^{20}} $ molecules.
The molecules are $ = 3.01 \times {10^{20}} $ molecules.
Additional Information:
The two important concepts of science are the mole and Avogadro ’s number that provides a link between the different properties of separate atoms. The number of atoms or other particles in a mole is the same for all the substances. Moles in general is related to the elements mass. We can say that one mole of carbon $ - 12 $ atoms have $ 6.02214076 \times {10^{23}} $ atoms and a mass of 12gram. In comparison to one mole of chlorine, by definition, of the same number of atoms as carbon $ - 12 $ ,but it has a mass of $ 35.453 $ u. When we are given volume, we can convert it into a number of moles present. This helps us to calculate the number of moles. one mole of ideal gas occupies $ 22.4litres $ of volume at STP. STP is the standard reference point of temperature $ (273.15k{\text{ or }}{{\text{0}}^ \circ }C) $ and pressure $ (1atm{\text{ or 1}}{{\text{0}}^5}pascals) $ used when measuring gas.
Note :
The answer will always be in SI unit or in the form the question is asking for. So it is important to convert the units properly according to the question. Focus on whether a question asks for STP or something else. Sometimes the answer is wrong due to incorrect methods applied.
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