Question

From $200mg$ of ${CO}_2$, ${10}^{21}$ molecules are removed the number of ${CO}_2$ molecules left are:

• A. $2.98\times {10}^{-3}$
• B. $1.66\times {10}^{-3}$
• C. $4.54\times {10}^{-2}$
• D. $1.66\times {10}^{-2}$

Answer 1

Answer: (A)

$2.98\times {10}^{-3}$

$200mg$ of ${CO}_2\equiv \frac{200\times {10}^{-3}}{44}$moles of ${CO}_2$

$\equiv 4.54\times {10}^{-3}$ moles of ${CO}_2$

$1$ mole of ${CO}_2$ contains $6.022\times {10}^{23}$ molecules $4.54\times {10}^{-3}$ moles of ${CO}_2$ will contain $2.737\times {10}^{21}$ molecules

No of molecules left after removing ${10}^{21}$ molecules $=\frac{(2.737\times {10}^{21}-{10}^{21})}{6.022\times {10}^{23}}$

$=2.88\times {10}^{-3}$