
The number of sodium atoms in \[2{\text{ }}moles\] of sodium ferrocyanide is:
\[
A.\;12{\text{ }}X{\text{ }}{10^{23}} \\
B.\;26{\text{ }}X{\text{ }}{10^{23}} \\
C.\;34{\text{ }}X{\text{ }}{10^{23}} \\
D.\;48{\text{ }}X{\text{ }}{10^{23}} \\
\]
Answer
485.1k+ views
Hint: We must remember the molecular formula for Sodium ferrocyanide is \[N{a_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]\]. \[1{\text{ }}mole\] of the compound is having \[4{\text{ }}moles\]of\[Na\]. Generally one mole of a substance is equal to \[6.022 \times {10^{23}}\]. The number \[6.022 \times {10^{23}}\] is known as Avogadro number.
Complete step by step answer:
Let’s start with writing the molecular formula for sodium ferrocyanide. As the name suggests it is a complex compound having 2 components which are metal and a complex. The complex consists of iron \[\left( {Fe} \right)\] and cyanide \[\left( {CN} \right)\]. As we can see the iron is in a ferrous state means having 2+ oxidation states. Also 6 \[CN\] are connected with iron so the overall oxidation state of the complex is -4. Hence, 4 \[N{a^ + }\] will be attached with the complex and the compounds molecular formula will be \[N{a_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]\].
\[1{\text{ }}mole\] of \[N{a_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]\] is having \[4{\text{ }}moles\] of sodium, and one mole of compound is having \[6.022{\text{ }} \times {\text{ }}{10^{23}}\] times the atom. So, \[4{\text{ }}moles\] of sodium will be having \[4{\text{ }} \times {\text{ }}6.022{\text{ }} \times {\text{ }}{10^{23}}\] atoms.
Similarly, in case of \[2{\text{ }}moles\], 2 times the atom in \[1{\text{ }}mole\] will be present so, \[8{\text{ }}moles\] of sodium will be there and hence \[8{\text{ }} \times {\text{ }}6.022{\text{ }} \times {\text{ }}{10^{23}}\] atoms which will be equal to \[48{\text{ }} \times {\text{ }}{10^{23}}\] atoms.
So, the answer to this question is D. \[48{\text{ }} \times {\text{ }}{10^{23}}\] atoms.
Note: We must know that the Avogadro’s number is being a boon for scientists as it helps in calculating, discussing and comparing very high numbers of atomic and subatomic particles. Avogadro’s number is \[6.022{\text{ }} \times {\text{ }}{10^{23}}\]. Avogadro’s number becomes very useful because in everyday life the substances contain a large number of atoms and molecules.
Complete step by step answer:
Let’s start with writing the molecular formula for sodium ferrocyanide. As the name suggests it is a complex compound having 2 components which are metal and a complex. The complex consists of iron \[\left( {Fe} \right)\] and cyanide \[\left( {CN} \right)\]. As we can see the iron is in a ferrous state means having 2+ oxidation states. Also 6 \[CN\] are connected with iron so the overall oxidation state of the complex is -4. Hence, 4 \[N{a^ + }\] will be attached with the complex and the compounds molecular formula will be \[N{a_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]\].
\[1{\text{ }}mole\] of \[N{a_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]\] is having \[4{\text{ }}moles\] of sodium, and one mole of compound is having \[6.022{\text{ }} \times {\text{ }}{10^{23}}\] times the atom. So, \[4{\text{ }}moles\] of sodium will be having \[4{\text{ }} \times {\text{ }}6.022{\text{ }} \times {\text{ }}{10^{23}}\] atoms.
Similarly, in case of \[2{\text{ }}moles\], 2 times the atom in \[1{\text{ }}mole\] will be present so, \[8{\text{ }}moles\] of sodium will be there and hence \[8{\text{ }} \times {\text{ }}6.022{\text{ }} \times {\text{ }}{10^{23}}\] atoms which will be equal to \[48{\text{ }} \times {\text{ }}{10^{23}}\] atoms.
So, the answer to this question is D. \[48{\text{ }} \times {\text{ }}{10^{23}}\] atoms.
Note: We must know that the Avogadro’s number is being a boon for scientists as it helps in calculating, discussing and comparing very high numbers of atomic and subatomic particles. Avogadro’s number is \[6.022{\text{ }} \times {\text{ }}{10^{23}}\]. Avogadro’s number becomes very useful because in everyday life the substances contain a large number of atoms and molecules.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
