Calculate the number of atoms of each element present in $122.5g$of $KCI{O_3}$.
Answer
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Hint: $1$ mole of a substance contains $6.022\, \times \,{10^{23}}$amount of that substance, which is Avogadro's number. A mole is the amount of substance which has mass equal to the gram atomic mass or gram molecular mass.
Formula used:
Molecular mass$ = $sum of atomic masses of all atoms present in that molecule.
Number of moles in a substance $ = $$\dfrac{{Given\,\,mass\,of\,the\,subs\tan ce}}{{Molecular\,mass}}$
$1$ mole of an atom = $6.022\, \times \,{10^{23}}$atoms
Complete step by step answer:
$1$ mole of a molecule is the amount of substance which has mass equal to the gram molecular mass. A gram molecular mass of a molecule is the sum of the atomic masses of all the atoms present in that molecule. $1$ mole of each atom is present in $1$ mole of a molecule.
So, $1$ mole of a $KCl{O_3}$ molecule contains $1$ mole of $K$ atom, $1$ mole of $Cl$ atom, and $3$ moles of $O$ atom.
Thus, the molecular mass of $KCl{O_3}$\[ = \,1 \times 39 + 1 \times 35.5 + 3 \times 16 = 122.5\,g/mol\]
By the mole concept,
Number of moles in a substance $ = $$\dfrac{{Given\,\,mass\,of\,the\,subs\tan ce}}{{Molecular\,mass}}$
Here, the given mass of $KCI{O_3}$$ = $$122.5g$
Thus, the number of moles in $122.5g $$KCI{O_3}$$ = $$\frac{{122.5g}}{{122.5g/mol}} = 1mole$
Therefore,
$1$ mole of $KCI{O_3}$ contains $1$ mole of $K$ atom = $6.022\, \times \,{10^{23}}$$K$atoms
$1$ mole of $KCI{O_3}$ contains $1$ mole of $Cl$ atom = $6.022\, \times \,{10^{23}}$$Cl$atoms
$1$ mole of $KCI{O_3}$ contains $3$ moles of $O$ atom = $3 \times 6.022\, \times \,{10^{23}}$$O$atoms$ = 18.066 \times {10^{23}}$$O$atoms
Note:
To calculate the molecular mass it is necessary to know the atomic masses of elements. There is 1 mole of each atom present in 1 mole of a molecule. Since, the given mass is equal to the molecular mass of $KCI{O_3}$, it can be easily interpreted that $6.022\, \times \,{10^{23}}$number of each atoms are present in $KCI{O_3}$. But, if the given mass is not equal to the molecular mass, the number of moles of the molecule present in the given mass should be calculated using the above formula, and for such a situation the number of atoms of each element will change.
Formula used:
Molecular mass$ = $sum of atomic masses of all atoms present in that molecule.
Number of moles in a substance $ = $$\dfrac{{Given\,\,mass\,of\,the\,subs\tan ce}}{{Molecular\,mass}}$
$1$ mole of an atom = $6.022\, \times \,{10^{23}}$atoms
Complete step by step answer:
$1$ mole of a molecule is the amount of substance which has mass equal to the gram molecular mass. A gram molecular mass of a molecule is the sum of the atomic masses of all the atoms present in that molecule. $1$ mole of each atom is present in $1$ mole of a molecule.
So, $1$ mole of a $KCl{O_3}$ molecule contains $1$ mole of $K$ atom, $1$ mole of $Cl$ atom, and $3$ moles of $O$ atom.
Thus, the molecular mass of $KCl{O_3}$\[ = \,1 \times 39 + 1 \times 35.5 + 3 \times 16 = 122.5\,g/mol\]
By the mole concept,
Number of moles in a substance $ = $$\dfrac{{Given\,\,mass\,of\,the\,subs\tan ce}}{{Molecular\,mass}}$
Here, the given mass of $KCI{O_3}$$ = $$122.5g$
Thus, the number of moles in $122.5g $$KCI{O_3}$$ = $$\frac{{122.5g}}{{122.5g/mol}} = 1mole$
Therefore,
$1$ mole of $KCI{O_3}$ contains $1$ mole of $K$ atom = $6.022\, \times \,{10^{23}}$$K$atoms
$1$ mole of $KCI{O_3}$ contains $1$ mole of $Cl$ atom = $6.022\, \times \,{10^{23}}$$Cl$atoms
$1$ mole of $KCI{O_3}$ contains $3$ moles of $O$ atom = $3 \times 6.022\, \times \,{10^{23}}$$O$atoms$ = 18.066 \times {10^{23}}$$O$atoms
Note:
To calculate the molecular mass it is necessary to know the atomic masses of elements. There is 1 mole of each atom present in 1 mole of a molecule. Since, the given mass is equal to the molecular mass of $KCI{O_3}$, it can be easily interpreted that $6.022\, \times \,{10^{23}}$number of each atoms are present in $KCI{O_3}$. But, if the given mass is not equal to the molecular mass, the number of moles of the molecule present in the given mass should be calculated using the above formula, and for such a situation the number of atoms of each element will change.
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