How many molecules are there in 2.3 grams of $N{{H}_{4}}S{{O}_{2}}$?
Answer
552k+ views
Hint: To solve this question, we first need to understand what is a mole. Mole is the SI unit of measurement and is used to determine the amount of a substance. A mole of any substance has exactly $6.022\times {{10}^{23}}$ particles which can be ions, atoms, electrons, or molecules.
Complete answer:
We know that if one mole of a substance is present, it has exactly the Avogadro number $({{N}_{A}})$ of particles.
${{N}_{A}}=6.022\times {{10}^{23}}$
So, the number of particles or molecules in n moles of a substance will be
\[N=n\times {{N}_{A}}\] (eq (i))
Where N is the number of molecules in the substance.
Now, the molecular mass of one mole of a compound is equal to the sum of atomic masses of all the elements present in the molecule.
So, the molecular mass of $N{{H}_{4}}S{{O}_{2}}$ is
\[\begin{align}
& {{M}_{N{{H}_{4}}S{{O}_{2}}}}={{M}_{N}}+4\times {{M}_{H}}+{{M}_{s}}+2\times {{M}_{O}} \\
& {{M}_{N{{H}_{4}}S{{O}_{2}}}}=14.0067+4\times 1.00784+32.065+2\times 15.999 \\
& {{M}_{N{{H}_{4}}S{{O}_{2}}}}=82.101g/mol \\
\end{align}\]
It is given to us that the mass of the sample substance is 2.3 grams.
So, the number of moles of the substance in the sample is
\[\begin{align}
& n=\dfrac{2.3g}{82.101g/mol} \\
& n\cong 0.028mol \\
\end{align}\]
By substituting the number of moles calculated above in eq(i), we get
\[\begin{align}
& N=n\times 6.022\times {{10}^{23}} \\
& N=0.028\times 6.022\times {{10}^{23}} \\
& N=1.686\times {{10}^{22}} \\
\end{align}\]
Hence, there are approximately $1.686\times {{10}^{22}}$ molecules in 2.3 grams of $N{{H}_{4}}S{{O}_{2}}$.
Note:
It should be noted that the number of particles in a mole of a given substance does not depend on the type or nature of a substance and the mass of all the particles ($6.022\times {{10}^{23}}$) in one mole of a substance is equivalent to the molar mass of that compound.
The mass of one mole of a substance is known as the molar mass of that substance. Its SI base unit is kg/mol but it is usually expressed in g/mol. It is a bulk property of a substance, not a molecular property.
Complete answer:
We know that if one mole of a substance is present, it has exactly the Avogadro number $({{N}_{A}})$ of particles.
${{N}_{A}}=6.022\times {{10}^{23}}$
So, the number of particles or molecules in n moles of a substance will be
\[N=n\times {{N}_{A}}\] (eq (i))
Where N is the number of molecules in the substance.
Now, the molecular mass of one mole of a compound is equal to the sum of atomic masses of all the elements present in the molecule.
So, the molecular mass of $N{{H}_{4}}S{{O}_{2}}$ is
\[\begin{align}
& {{M}_{N{{H}_{4}}S{{O}_{2}}}}={{M}_{N}}+4\times {{M}_{H}}+{{M}_{s}}+2\times {{M}_{O}} \\
& {{M}_{N{{H}_{4}}S{{O}_{2}}}}=14.0067+4\times 1.00784+32.065+2\times 15.999 \\
& {{M}_{N{{H}_{4}}S{{O}_{2}}}}=82.101g/mol \\
\end{align}\]
It is given to us that the mass of the sample substance is 2.3 grams.
So, the number of moles of the substance in the sample is
\[\begin{align}
& n=\dfrac{2.3g}{82.101g/mol} \\
& n\cong 0.028mol \\
\end{align}\]
By substituting the number of moles calculated above in eq(i), we get
\[\begin{align}
& N=n\times 6.022\times {{10}^{23}} \\
& N=0.028\times 6.022\times {{10}^{23}} \\
& N=1.686\times {{10}^{22}} \\
\end{align}\]
Hence, there are approximately $1.686\times {{10}^{22}}$ molecules in 2.3 grams of $N{{H}_{4}}S{{O}_{2}}$.
Note:
It should be noted that the number of particles in a mole of a given substance does not depend on the type or nature of a substance and the mass of all the particles ($6.022\times {{10}^{23}}$) in one mole of a substance is equivalent to the molar mass of that compound.
The mass of one mole of a substance is known as the molar mass of that substance. Its SI base unit is kg/mol but it is usually expressed in g/mol. It is a bulk property of a substance, not a molecular property.
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