Question

The number of iodine atom present in $50ml$ of a $0.1M$ $KI$ solution is

• A. $6\times {10}^{23}$
• B. $12\times {10}^{23}$
• C. $3\times {10}^{21}$
• D. $6\times {10}^{22}$

Answer 1

The correct option is C $3\times {10}^{21}$

Molarity(M) = $\frac{n}{V\left(ml\right)}\times 1000$

$\therefore 0.1=\frac{n}{50}\times 1000$

$\therefore$ no. of moles of KI = $5\times {10}^{-3}$

1 mole of KI contains = $6.02\times {10}^{23}$ atoms

$\therefore$ $5\times {10}^{-3}$ contains = $5\times {10}^{-3}\times 6.02\times {10}^{23}$ = $3.011\times {10}^{21}$