Question

The number of iodine atoms present in $1c{m}^3$ of its 0.1 M solution is

• A. $12.04\times 10^{23}$
• B. $6.02\times 10^{22}$
• C. $12.04\times 10^{19}$
• D. $6.02\times 10^{20}$

Answer 1

Answer: (C)

$\text{1000}{\text{cm}}^{\text{3}}\text{of}\text{0}\text{.1}\text{M}\text{solution}\equiv \text{0}\text{.1}\text{mol}{\text{I}}_{\text{2}}$

So $1{\text{cm}}^{\text{3}}\text{of}\text{solution}\equiv 0.1\times 10^{-3}\text{mol}{\text{I}}_{\text{2}}$

Hence $\text{1}\text{mol}{\text{I}}_{\text{2}}$ contains $=6.02\times 10^{23}{\text{I}}_{\text{2}}$ molecules

So $0.1\times 10^{-3}\text{mol}{\text{I}}_{\text{2}}$ contains

$=6.02\times 10^{23}\times 0.1\times 10^{-3}{\text{I}}_{\text{2}}$ molecules

$=12.04\times 10^{19}{\text{I}}_{\text{2}}$ atoms.