Question

When x molecules are removed from 200 mg of ${N_2}O$, $2.89 \times {10^{ - 3}}$ moles of ${N_2}O$ are left in x will be
A. ${10^{20}}$ molecules
B. ${10^{10}}$ molecules
C. 21 molecules
D. ${10^{21}}$ molecules

Answer 1

Hint: - For solving these types of questions we must remember the molecular mass of the given compound. Also, we must know the definition and value of mole.

Complete answer:
Molecular mass of ${N_2}O$ = 44 and weight of ${N_2}O$= 200mg = 0.2g
Moles of ${N_2}O$ present = $\dfrac{{0.2}}{{44}}$= $4.55 \times {10^{ - 3}}$
Let moles of N₂O removed be $x$, therefore moles of ${N_2}O$ remained = $2.89 \times {10^{ - 3}}$moles
Thus $4.55 \times {10^{ - 3}}$- $x$= $2.89 \times {10^{ - 3}}$
$x$=$4.55 \times {10^{ - 3}}$- $2.89 \times {10^{ - 3}}$
$x$=$1.65 \times {10^{ - 3}}$mole
As we know in 1 mole there are $6.022 \times {10^{23}}$ molecules
Therefore, $1.65 \times {10^{ - 3}}$= $6.022 \times {10^{23}}$$ \times $ $1.65 \times {10^{ - 3}}$
$9.97 \times {10^{20}}$molecules
${10^{21}}$ molecules

Therefore, option D is the correct answer and when $x$ molecules are removed from 200 mg of${N_2}O$ moles of ${N_2}O$ left will be ${10^{21}}$ molecules.

Note: - In this question we saw the whole question was based on simple calculation but the values are very important. We must remember the value like in 1 mole there are $6.022 \times {10^{23}}$ molecules. This is very important.