Question

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

• A. $6.02\times {10}^{23}$
• B. $6.02\times {10}^{22}$
• C. $6.02\times {10}^{19}$
• D. $1.204\times {10}^{20}$

Answer 1

Answer: (C)

$6.02\times {10}^{19}$

No. of iodine atoms in ${1cm}^3$ of $0-1M$ solution

$0.1M$ solution means

$0.1$moles is dissolved in $1L$ of solution

$1L\to 0.1moles$

Given volume=$1c{m}^3={10}^{-3}L$

Therefore by unitary method

${10}^{-3}L\to {10}^{-3}\times 0.1moles\to {10}^{-4}moles$

$1$mole contains$\to 6.022\times {10}^{23}$atoms

${10}^{-4}$mol contains$\to {10}^{-4}\times 6.022\times {10}^{23}\to 6.022\times {10}^{19}atoms$

The number of iodine atoms present in ${1cm}^3$ of $0.1M$ solution is $6.022\times {10}^{19}$atoms