Question

If ${10}^{20}$ molecules are removed from $3.4$ mg of $N{H}_3$, then the number of moles of $N{H}_3$ left is:

• A. $1.7\times {10}^{-5}$
• B. $3.3\times {10}^{-5}$
• C. $1\times {10}^{-4}$
• D. $8.5\times {10}^{-3}$

Answer 1

The correct option is B $3.3\times {10}^{-5}$

17 g of $N{H}_3$ has $6.023\times {10}^{23}$ molecules of $N{H}_3$

$⟹17⟶6.023\times {10}^{23}$

$3.4\times {10}^{-3}g⟹\left(x\right)$

$x=\frac{34}{17}\times 6.023\times {10}^{20}=1.2\times {10}^{20}$ molecules

When ${10}^{20}$ are removed, it has $(1.2-1)\times {10}^{20}=2\times {10}^{19}$

No. of moles left = $\frac{2}{6.02}\times \frac{{10}^{19}}{{10}^{23}}=\frac{1}{3}\times {10}^{-4}=3.3\times {10}^{-5}$