Question

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

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

Answer 1

The correct option is D ${10}^{21}$ molecules.

Molar mass of $N_{2}O=449$

No. of moles of $N_{2}O$ in $449=1$

No. of moles of $N_{3}O$ in $1g$ of $N_{2}O=\frac{1}{44}$

No. of molecules of $N_{2}O$ in $200mg(200\times {10}^3)=200\times {10}^{-3}/44=4.545\times {10}^{-3}$ moles.

No. of moles of $N_{2}O$ lost = $2.89\times {10}^{-3}$

Total no. of moles = $4.545\times {10}^{-3}$ moles

No. of moles removed = $4.545\times {10}^{-3}-2.89\times {10}^{-3}=1.8\times {10}^{-3}$ moles.

$1$ moles of $N_{2}O=6.022\times {10}^{23}$ moles of $N_{2}O$

$1.655\times {10}^{-3}$ moles of $N_{2}O$

$=1.65\times {10}^{-3}\times 6.022\times {10}^{23}$ moles of $N_{2}O$

$=9.96\times {10}^{20}$ moles of $N_{2}O$

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