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How do you solve for\[{{V}_{2}}\] , using Avogadro’s law when moles of a gas are added \[{{n}_{2}}\]. A sample containing 7.20g oxygen gas has a volume of 35.0L. pressure and temperature remain constant. What is the new volume if 0.800 moles of oxygen gas is added?

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: Mole is defined as the basic unit for measuring the amount of substance. The mole is defined as the unit which contains \[6.023\times {{10}^{23}}\]elementary entities of the substance. The \[6.023\times {{10}^{23}}\]is the avogadro number. The 1 mole of the substance contains the amount of substance that is present in 12g of the carbon 12 atom.

Complete step by step solution:
So to calculate the \[{{V}_{2}}\]we need to calculate many things before it.
So first convert the given grams of oxygen into moles by dividing the given mass by its molecular weight of oxygen. so the formula which will be used is following:
Moles=\[\dfrac{\text{mass}}{\text{molecular mass}}\]
The values in the question given is
Mass $= 7.20g$
Molecular mass $= 2$(atomic mass of oxygen)
Atomic mass of oxygen $= 16$
So the molecular weight will be $= 2(16)= 32g/mol$
Now substituting the values in formula we get,
Moles \[=\dfrac{7.20}{32}\]
Moles $= 0.225$
So now the question says that 0.800 moles of oxygen are further added to the original one that is the
\[{{n}_{2}}\] $= 0.225+0.800= 1.025 $mole
Now the formula which we are going to us is the following:
\[{{V}_{1}}{{n}_{2}}={{V}_{2}}{{n}_{1}}\]
The values given to us are
\[{{V}_{1}}\]= 35.0L
\[{{n}_{2}}\]= 1.025 moles
\[{{n}_{1}}\]= 0.225 moles
Now substituting the values,
\[{{V}_{2}}=35.0\times \dfrac{1.025}{0.225}=159L\]
The volume after the 0.800 moles of oxygen is added is 159L.

Note: in 1 mole of the substance the number of atoms or molecules is present is equal to \[6.023\times {{10}^{23}}\]. The avogadro number helps in creating the relationship between the physical constants. The atomic level substances are measured in the atomic mass unit. The one amu is equal to \[1.66\times {{10}^{-24}}grams\].
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