
The incorrect statement for 14g CO is:
(A) It occupies 2.24L at NTP.
(B) It corresponds to 0.5 moles of CO.
(C) It corresponds to the same moles of CO and ${{N}_{2}}$.
(D) It corresponds to $3.01\,\times \,{{10}^{23}}$ molecules of CO.
Answer
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Hint: CO also known as carbon monoxide is a harmful toxic gas. It is one of the major constituents of pollutant gases. It is released if the fuel is burnt in insufficient amounts of oxygen. CO also causes carbon monoxide poisoning in humans if inhaled in excess.
The approach to answer this question is by calculating the moles of CO present in 14g.
Complete answer:
Let’s look at the solution to this question:
First, we will calculate the number of moles in 14g of CO.
Given mass of CO = 14g
Molar mass of CO = 28g
\[Moles\,of\,CO\,=\,\dfrac{Given\,Mass}{Molar\,mass}\]
\[Moles\,of\,CO\,=\,\dfrac{14}{28}\,=\,0.5\]
So, the number of moles present in 14g of CO is 0.5
So, the given statement in option (B) is correct.
Now, we will calculate the moles in 14g of ${{N}_{2}}$
Given mass of ${{N}_{2}}$= 14g
Molar mass of ${{N}_{2}}$= 28g
\[Moles\,of\,{{N}_{2\,}}\,=\,\dfrac{Given\,Mass}{Molar\,mass}\]
\[Moles\,of\,{{N}_{2\,}}\,=\,\dfrac{14}{28}\,=0.5\]
So, the number of moles present in CO and ${{N}_{2}}$ is the same.
Hence, the given statement in option (C) is correct.
Now, we will calculate the number of molecules present in 14 grams of ${{N}_{2}}$
\[Number\,of\,molecules\,=\,Moles\times Avogadro\,number\]
\[Number\,of\,molecules\,of\,{{N}_{2}}\,=\,0.5\times 6.02\times {{10}^{23}}\,=\,3.01\times {{10}^{23}}\]
So, the number of molecules present in 14g of ${{N}_{2}}$ is $3.01\times {{10}^{23}}$.
So, the statement given in option (D) is correct.
Now, we know that 1 mole of a gas occupies 22.4L at NTP,
Therefore, 0.5 moles will occupy $0.5\times 22.4\,=\,11.2L$
So, the statement given in option (A) is incorrect.
Hence, the answer for the given question is option (A).
Note: At NTP the volume occupied by an ideal gas is 22.4L. This is true for all gases. This happens because the volume of a gas molecule is considered negligible according to the kinetic theory of gases.
The approach to answer this question is by calculating the moles of CO present in 14g.
Complete answer:
Let’s look at the solution to this question:
First, we will calculate the number of moles in 14g of CO.
Given mass of CO = 14g
Molar mass of CO = 28g
\[Moles\,of\,CO\,=\,\dfrac{Given\,Mass}{Molar\,mass}\]
\[Moles\,of\,CO\,=\,\dfrac{14}{28}\,=\,0.5\]
So, the number of moles present in 14g of CO is 0.5
So, the given statement in option (B) is correct.
Now, we will calculate the moles in 14g of ${{N}_{2}}$
Given mass of ${{N}_{2}}$= 14g
Molar mass of ${{N}_{2}}$= 28g
\[Moles\,of\,{{N}_{2\,}}\,=\,\dfrac{Given\,Mass}{Molar\,mass}\]
\[Moles\,of\,{{N}_{2\,}}\,=\,\dfrac{14}{28}\,=0.5\]
So, the number of moles present in CO and ${{N}_{2}}$ is the same.
Hence, the given statement in option (C) is correct.
Now, we will calculate the number of molecules present in 14 grams of ${{N}_{2}}$
\[Number\,of\,molecules\,=\,Moles\times Avogadro\,number\]
\[Number\,of\,molecules\,of\,{{N}_{2}}\,=\,0.5\times 6.02\times {{10}^{23}}\,=\,3.01\times {{10}^{23}}\]
So, the number of molecules present in 14g of ${{N}_{2}}$ is $3.01\times {{10}^{23}}$.
So, the statement given in option (D) is correct.
Now, we know that 1 mole of a gas occupies 22.4L at NTP,
Therefore, 0.5 moles will occupy $0.5\times 22.4\,=\,11.2L$
So, the statement given in option (A) is incorrect.
Hence, the answer for the given question is option (A).
Note: At NTP the volume occupied by an ideal gas is 22.4L. This is true for all gases. This happens because the volume of a gas molecule is considered negligible according to the kinetic theory of gases.
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