Question

What is the ratio of the number of atoms of elements potassium, chlorine, and oxygen respectively present in $245g$ of $KCl{O}_3$?

• A. $2:2:3$
• B. $1:1:3$
• C. $39:36:48$
• D. $1:2:6$

Answer 1

The correct option is B $1:1:3$

$KCl{O}_3⟶$ molecular weight = $122.5gmo{l}^{-1}$

There is $1$ mole of potassium in $1$ mole of $KCl{O}_3=1\times 39=39g$

There is $1$ mole of $Cl=1\times 35.5=35.5g$

and $3$ mole of oxygen $=3\times 16=48g$

Moles of $KCl{O}_3=\frac{Givenweight}{molecularweight}$

$=\frac{245g}{122.5gmo{l}^{-1}}$

Moles of $KCl{O}_3=2$

$\therefore$ Moles of Potassium $=2$

Moles of Chlorine $=2$

Moles of Oxygen $=6$

(1) Number of atoms of $K$ = Moles $\times$ Avogadro's number

$=2\times 6.022\times {10}^{23}$

$=12.044\times {10}^{23}$

(2) Number of atoms of $Cl$ = $2\times 6.022\times {10}^{23}$

$=12.044\times {10}^{23}$

(3) Number of atoms of Oxygen = $6\times 6.022\times {10}^{23}$

$=36.132\times {10}^{23}$

Ratio of number of atoms of $K,Cl$ & $O=1:1:3$ .