
\[16g\] of oxygen has same number of molecules as in
A. \[16g\]of \[CO\]
B. \[28g\]of \[{N_2}\]
C. \[14g\]of \[{N_2}\]
D. \[1.0g\] of \[{H_2}\]
Answer
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Hint: In this we need to calculate the number of molecules which can be calculated as mass/ molar mass multiplied by the Avogadro number. Further look which are matching to that of the oxygen atom that would be the answer.
Complete step-by-step answer:Here we need to calculate the same number of molecules as present in the \[16g\] of oxygen atom .so it can be calculated as
\[\dfrac{{16}}{{32}} \times {{\rm N}_A} = \dfrac{{{{\rm N}_A}}}{2}\]
Further we need to calculate the number of molecules in each of the given option
So considering the first option that is sixteen grams of carbon monoxide and this can be further calculated as
\[CO = \dfrac{{16}}{{28}} \times {{\rm N}_A} = \dfrac{{4{{\rm N}_A}}}{7}\]
So it is not matching with the number of molecules of the oxygen atom.so we will go for the second option that is the twenty eight grams of nitrogen.so its number of molecules will be calculated as
\[{N_2} = \dfrac{{28}}{{28}} \times {{\rm N}_A} = {{\rm N}_A}\]
Here it also not matching the number of molecules of oxygen so we will look for the other options.
So, the third option is fourteen grams of nitrogen
\[{N_2} = \dfrac{{14}}{{28}} \times {{\rm N}_A} = \dfrac{{{{\rm N}_A}}}{2}\]
So it is matching the numHere we need to calculate the number of molecules which can be calculated as mass/ molar mass multiplied by the Avogadro number.
ber of molecules of the oxygen . but to check that whether other options are correct or not we will calculate their number of molecules also
Lastly we have one grams of hydrogen
\[{H_2} = \dfrac{1}{2} \times {{\rm N}_A} = \dfrac{{{{\rm N}_A}}}{2}\]
It is also matching the number of molecules of oxygen atoms .
So we concluded here that option C and D are correct.
Note:We should further note here that the mass of the mole of the substance is equal to the substances molecular weight.further the mole is the amount of a substance of a system that contains many entities.
Complete step-by-step answer:Here we need to calculate the same number of molecules as present in the \[16g\] of oxygen atom .so it can be calculated as
\[\dfrac{{16}}{{32}} \times {{\rm N}_A} = \dfrac{{{{\rm N}_A}}}{2}\]
Further we need to calculate the number of molecules in each of the given option
So considering the first option that is sixteen grams of carbon monoxide and this can be further calculated as
\[CO = \dfrac{{16}}{{28}} \times {{\rm N}_A} = \dfrac{{4{{\rm N}_A}}}{7}\]
So it is not matching with the number of molecules of the oxygen atom.so we will go for the second option that is the twenty eight grams of nitrogen.so its number of molecules will be calculated as
\[{N_2} = \dfrac{{28}}{{28}} \times {{\rm N}_A} = {{\rm N}_A}\]
Here it also not matching the number of molecules of oxygen so we will look for the other options.
So, the third option is fourteen grams of nitrogen
\[{N_2} = \dfrac{{14}}{{28}} \times {{\rm N}_A} = \dfrac{{{{\rm N}_A}}}{2}\]
So it is matching the numHere we need to calculate the number of molecules which can be calculated as mass/ molar mass multiplied by the Avogadro number.
ber of molecules of the oxygen . but to check that whether other options are correct or not we will calculate their number of molecules also
Lastly we have one grams of hydrogen
\[{H_2} = \dfrac{1}{2} \times {{\rm N}_A} = \dfrac{{{{\rm N}_A}}}{2}\]
It is also matching the number of molecules of oxygen atoms .
So we concluded here that option C and D are correct.
Note:We should further note here that the mass of the mole of the substance is equal to the substances molecular weight.further the mole is the amount of a substance of a system that contains many entities.
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