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Question: When x molecules are removed from 200 mg of \({N_2}O\), \(2.89 \times {10^{ - 3}}\) moles of \({N_2}...

When x molecules are removed from 200 mg of N2O{N_2}O, 2.89×10−32.89 \times {10^{ - 3}} moles of N2O{N_2}O are left in x will be
A. 1020{10^{20}} molecules
B. 1010{10^{10}} molecules
C. 21 molecules
D. 1021{10^{21}} molecules

Explanation

Solution

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 N2O{N_2}O = 44 and weight of N2O{N_2}O= 200mg = 0.2g
Moles of N2O{N_2}O present = 0.244\dfrac{{0.2}}{{44}}= 4.55×10−34.55 \times {10^{ - 3}}
Let moles of N₂O removed be xx, therefore moles of N2O{N_2}O remained = 2.89×10−32.89 \times {10^{ - 3}}moles
Thus 4.55×10−34.55 \times {10^{ - 3}}- xx= 2.89×10−32.89 \times {10^{ - 3}}
xx=4.55×10−34.55 \times {10^{ - 3}}- 2.89×10−32.89 \times {10^{ - 3}}
xx=1.65×10−31.65 \times {10^{ - 3}}mole
As we know in 1 mole there are 6.022×10236.022 \times {10^{23}} molecules
Therefore, 1.65×10−31.65 \times {10^{ - 3}}= 6.022×10236.022 \times {10^{23}} ×\times 1.65×10−31.65 \times {10^{ - 3}}
9.97×10209.97 \times {10^{20}}molecules
1021{10^{21}} molecules

Therefore, option D is the correct answer and when xx molecules are removed from 200 mg ofN2O{N_2}O moles of N2O{N_2}O left will be 1021{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 6.022×10236.022 \times {10^{23}} molecules. This is very important.