
How many moles of sodium carbonate are present in 6.80 grams of sodium carbonate?
Answer
496.8k+ views
Hint:The mole concept is used to determine the amount of the substance which is obtained by taking the ratio of the mass of the substance to the molecular mass of the substance.
One mole of the substance contains the Avogadro’s numbers of the particles that are atoms, ions, or molecules of the substance.Avogadro’s number is represented as \[{{\text{N}}_{\text{A}}}\]. The value of Avogadro’s number is \[{\text{6}}{\text{.023 \times 1}}{{\text{0}}^{{\text{23}}}}\,{\text{molecules}}\].
The molecular formula of the sodium carbonate is \[{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}\] and molecular weight of the sodium carbonate is \[{\text{105}}{\text{.9888}}\,{\text{g/mol}}\].
Formula used: The moles of the substance calculated as follows:
\[{\text{moles = }}\dfrac{{{\text{weight}}}}{{{\text{Molecular}}\,{\text{weight}}}}\]
Complete step-by-step solution:Here, the molecular weight of the sodium carbonate given is \[{\text{105}}{\text{.9888}}\,{\text{g/mol}}\] and the weight of the sodium carbonate is 6.80 grams.
We have to determine moles of the sodium carbonate in 6.80 grams as follows:
\[{\text{moles = }}\dfrac{{{\text{weight}}}}{{{\text{Molecular}}\,{\text{weight}}}}\]
Substitute, \[{\text{105}}{\text{.9888}}\,{\text{g/mol}}\] for molecular weight and 6.80 grams for weight.
\[{\text{moles = }}\dfrac{{6.80\,{\text{g}}}}{{{\text{105}}{\text{.9888}}\,{\text{g/mol}}}}\]
\[\therefore {\text{moles = 0}}{\text{.064158}}\,{\text{mol}}\]
Thus, the moles of the sodium carbonate present in 6.80 grams are \[{\text{0}}{\text{.0642}}\,{\text{mol}}\].
Note:Molecular weight is the average mass of the molecules and it is deferrable from molecule to molecule. In the case of the atom, it is called atomic mass while in the case of the molecular mass it is calculated by taking the sum of the atomic masses of all the atoms in the molecule.Here, we have to use the molecular weight of the sodium carbonate. The SI unit of the molecular weight is gram per mole. Its unit itself indicates the ratio of the mass of the substance to the molar molecular mass of the substance.
One mole of the substance contains the Avogadro’s numbers of the particles that are atoms, ions, or molecules of the substance.Avogadro’s number is represented as \[{{\text{N}}_{\text{A}}}\]. The value of Avogadro’s number is \[{\text{6}}{\text{.023 \times 1}}{{\text{0}}^{{\text{23}}}}\,{\text{molecules}}\].
The molecular formula of the sodium carbonate is \[{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}\] and molecular weight of the sodium carbonate is \[{\text{105}}{\text{.9888}}\,{\text{g/mol}}\].
Formula used: The moles of the substance calculated as follows:
\[{\text{moles = }}\dfrac{{{\text{weight}}}}{{{\text{Molecular}}\,{\text{weight}}}}\]
Complete step-by-step solution:Here, the molecular weight of the sodium carbonate given is \[{\text{105}}{\text{.9888}}\,{\text{g/mol}}\] and the weight of the sodium carbonate is 6.80 grams.
We have to determine moles of the sodium carbonate in 6.80 grams as follows:
\[{\text{moles = }}\dfrac{{{\text{weight}}}}{{{\text{Molecular}}\,{\text{weight}}}}\]
Substitute, \[{\text{105}}{\text{.9888}}\,{\text{g/mol}}\] for molecular weight and 6.80 grams for weight.
\[{\text{moles = }}\dfrac{{6.80\,{\text{g}}}}{{{\text{105}}{\text{.9888}}\,{\text{g/mol}}}}\]
\[\therefore {\text{moles = 0}}{\text{.064158}}\,{\text{mol}}\]
Thus, the moles of the sodium carbonate present in 6.80 grams are \[{\text{0}}{\text{.0642}}\,{\text{mol}}\].
Note:Molecular weight is the average mass of the molecules and it is deferrable from molecule to molecule. In the case of the atom, it is called atomic mass while in the case of the molecular mass it is calculated by taking the sum of the atomic masses of all the atoms in the molecule.Here, we have to use the molecular weight of the sodium carbonate. The SI unit of the molecular weight is gram per mole. Its unit itself indicates the ratio of the mass of the substance to the molar molecular mass of the substance.
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