
Which of the following contains the least number of molecules $ ? $
(A) $ 3.2\,g\,S{O_2} $
(B) $ 3.4\,g\,N{H_3} $
(C) $ 1.6\,g\,C{H_4} $
(D) $ 3.2\,g\,S{O_2} $
Answer
435.9k+ views
Hint :To solve the above question, we must understand the mole concept. Mole (mol) is the measure of the amount of substance present in a material. A mole contains $ 6.022 \times {10^{23}} $ number of entities which can be atoms, molecules or ions. The number $ 6.022 \times {10^{23}} $ is known as Avogadro’s number $ ({N_A}) $ .
Complete Step By Step Answer:
From the hint given, we understood that a mole is the amount of a substance that contains $ 6.022 \times {10^{23}} $ particles. In order to find out which compound contains the least number of molecules, we must calculate the number of moles present in them.
Number of moles , $ n = \dfrac{{given\,mass\,(m)}}{{molar\,mass\,(M)}} $
Also, $ n = \dfrac{{number\,of\,molecules\,(N)}}{{Avogadro\,number\,({N_A})}} $
Therefore, number of molecules (N) $ = \dfrac{m}{M} \times {N_A} $
First, we have to calculate the molar masses of given compounds and then substitute it in the given equation. The molar mass of compounds are :
$C{O_2}\, = \,44g\,mo{l^{ - 1}} \\
N{H_3}\, = 17g\,mo{l^{ - 1}} \\
C{H_4}\, = 16g\,mo{l^{ - 1}} \\
S{O_2}\, = 64g\,mo{l^{ - 1}} \\ $
Number of molecules in $ C{O_2} $ $ = \dfrac{{4.4}}{{44}} \times {N_A} = 0.1{N_A} $
Number of molecules in $ N{H_{3\,}}\, = \dfrac{{3.4}}{{17}} \times {N_A} = 0.2{N_A} $
Number of molecules in $ C{H_4}\, = \dfrac{{1.6}}{{16}} \times {N_A} = 0.1{N_A} $
Number of molecules in $ S{O_2}\, = \dfrac{{3.2}}{{64}} \times {N_A} = 0.05{N_A} $
$ {N_A} = 6.022 \times {10^{23}} $ molecules
Therefore, the right option is (D) $ 3.2\,g\,S{O_2} $ . It contains the least number of molecules.
Note :
Molar mass (M) is the mass of one mole of a substance, or mass of $ 6.022 \times {10^{23}} $ number of entities. It is expressed in $ gmo{l^{ - 1}} $ .It is often used interchangeably with molecular mass. Molecular mass is defined as the mass of a single molecule. The unit of molecular mass is given in atomic mass units (amu).
Given, the compounds are present in equal quantity (equal grams), then the molecule having higher molar mass will contain the least number of molecules.
Complete Step By Step Answer:
From the hint given, we understood that a mole is the amount of a substance that contains $ 6.022 \times {10^{23}} $ particles. In order to find out which compound contains the least number of molecules, we must calculate the number of moles present in them.
Number of moles , $ n = \dfrac{{given\,mass\,(m)}}{{molar\,mass\,(M)}} $
Also, $ n = \dfrac{{number\,of\,molecules\,(N)}}{{Avogadro\,number\,({N_A})}} $
Therefore, number of molecules (N) $ = \dfrac{m}{M} \times {N_A} $
First, we have to calculate the molar masses of given compounds and then substitute it in the given equation. The molar mass of compounds are :
$C{O_2}\, = \,44g\,mo{l^{ - 1}} \\
N{H_3}\, = 17g\,mo{l^{ - 1}} \\
C{H_4}\, = 16g\,mo{l^{ - 1}} \\
S{O_2}\, = 64g\,mo{l^{ - 1}} \\ $
Number of molecules in $ C{O_2} $ $ = \dfrac{{4.4}}{{44}} \times {N_A} = 0.1{N_A} $
Number of molecules in $ N{H_{3\,}}\, = \dfrac{{3.4}}{{17}} \times {N_A} = 0.2{N_A} $
Number of molecules in $ C{H_4}\, = \dfrac{{1.6}}{{16}} \times {N_A} = 0.1{N_A} $
Number of molecules in $ S{O_2}\, = \dfrac{{3.2}}{{64}} \times {N_A} = 0.05{N_A} $
$ {N_A} = 6.022 \times {10^{23}} $ molecules
Therefore, the right option is (D) $ 3.2\,g\,S{O_2} $ . It contains the least number of molecules.
Note :
Molar mass (M) is the mass of one mole of a substance, or mass of $ 6.022 \times {10^{23}} $ number of entities. It is expressed in $ gmo{l^{ - 1}} $ .It is often used interchangeably with molecular mass. Molecular mass is defined as the mass of a single molecule. The unit of molecular mass is given in atomic mass units (amu).
Given, the compounds are present in equal quantity (equal grams), then the molecule having higher molar mass will contain the least number of molecules.
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