Question

From $440mg$ of ${CO}_2,{10}^{20}$ molecules of${CO}_2$ are removed. The total number of remaining atoms will be?

• A. $1.51\times {10}^{20}$
• B. $12.89\times {10}^{21}$
• C. $1.77\times {10}^{22}$
• D. $1.46\times {10}^{23}$

Answer 1

The correct option is C $1.77\times {10}^{22}$

Removed molecule $={10}^{20}$

$C{O}_2=440mg=0.44g$

Molecular weight of $C{O}_2=12+32=44$

Moles of $C{O}_2=\frac{0.44}{44}=0.01$

Number of molecule = $0.01\times 6.023\times {10}^{23}$

= $6.023\times {10}^{21}$

Remaining molecule = initial $-$ removed molecule

= $6.023\times {10}^{21}-0.1\times {10}^{21}$

= $5.923\times {10}^{21}$

Atom remaining = $3\times$ remaining molecule

= $3\times 5.923\times {10}^{21}=1.77\times {10}^{21}$

Hence, option(c) is the correct answer.