Question

When $x$ molecules are removed from $200mg{N}_{2}O,2.89\times {10}^{-3}$ moles of $N_{2}O$ are left. The value of $x$ will be:

• A. ${10}^{20}$ molecules
• B. ${10}^{10}$ molecules
• C. $21$ molecules
• D. ${10}^{21}$ molecules

Answer 1

Answer: (D)

${10}^{21}$ molecules

moles of $N_{2}O$ = $\frac{200\times {10}^{-3}}{44}=4.54\times {10}^{-3}$ moles

$\therefore$ $Molesremoved=4.54\times {10}^{-3}-2.89\times {10}^{-3}$

$=1.65\times {10}^{-3}$

$\therefore$ $Moleculesremoved=1.65\times {10}^{-3}\times 6.022\times {10}^{23}$

$=10\times {10}^{20}$

$={10}^{21}$ molecules.

Hence, option $D$ is correct.