Solveeit Logo
Login

Question

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

When xx molecules are removed from 200  mg  N2O200\;mg\;{N_2}O, 2.89×10−32.89 \times {10^{ - 3}} moles of N2O{N_2}O are left. The value of xx will be:
A. 1020molecules{10^{20}}{\rm{ molecules}}
B. 1010molecules{10^{10}}{\rm{ molecules}}
C. 21 molecules21 \,{\rm{ molecules}}
D. 1021molecules{10^{21}}{\rm{ molecules}}
Given:
- Number of removed molecules is assumed to be xx
- Initial mass of N2O{N_2}O was taken to be minitial=200  mg{m_{{\rm{initial}}}} = 200\;mg
- Final amount of N2O{N_2}O was found to be nfinal=2.89×10−3  mol{n_{{\rm{final}}}} = 2.89 \times {10^{ - 3}}\;mol

Explanation

Solution

We have the Avogadro constant that gives a relationship between the amount of matter and number of particles present in it which can be used to find the same.

Complete step by step solution:
We know that as per Avogadro’s law, 11 mole has been defined as the amount of matter which contains 6.02×10236.02 \times {10^{23}} particles of matter in it. From here, we get the Avogadro number or Avogadro constant as NA=6.02×1023{N_A} = 6.02 \times {10^{23}}. For 11 mole of HH , it would be 6.02×10236.02 \times {10^{23}} HH atoms, for 11 mole of H2{H_2} gas, it would be 6.02×10236.02 \times {10^{23}} H2{H_2}molecules.

We have one more concept related to the amount of the matter namely molar mass. It is the mass of 11 mole of matter and is constant for a given matter. For HH atoms it is 1  gmol−11\;gmo{l^{ - 1}} whereas for H2{H_2} gas it is 2  gmol−12\;gmo{l^{ - 1}}.

Here, we have N2O{N_2}O so we will consider its molecules. Let’s first convert the units of the given mass by using the conversion factor which is given below:
1  g1000  mg\dfrac{{1\;g}}{{1000\;mg}}

We will convert the units of minitial=200  mg{m_{{\rm{initial}}}} = 200\;mg by using the above conversion factor as follows:
minitial=200  mg×(1  g1000  mg) =0.2  g {m_{{\rm{initial}}}} = 200\;mg \times \left( {\dfrac{{1\;g}}{{1000\;mg}}} \right)\\\ = 0.2\;g

Now, we can write the conversion factor by using the molar mass of N2O{N_2}O which is 44.0  gmol−144.0\;gmo{l^{ - 1}} as follows:
1  mol44.0  g\dfrac{{1\;mol}}{{44.0\;g}}
We will calculate the initial amount by using the initial mass and the above conversion factor as follows:

ninitial=0.2  g×(1  mol44.0  g) =0.00454  mol =4.54×10−3  mol{n_{{\rm{initial}}}} = 0.2\;g \times \left( {\dfrac{{1\;mol}}{{44.0\;g}}} \right)\\\ = {\rm{0}}{\rm{.00454}}\;mol\\\ = {\rm{4}}{\rm{.54}} \times {\rm{1}}{{\rm{0}}^{ - 3}}\;mol

Now, we can calculate the amount of N2O{N_2}O that was removed initial amount by subtracting final one from the initial as follows:

nremoved=ninitial−nfinal =(4.54×10−3  mol)−(2.89×10−3  mol) =1.65×10−3  mol{n_{{\rm{removed}}}} = {n_{{\rm{initial}}}} - {n_{{\rm{final}}}}\\\ = \left( {{\rm{4}}{\rm{.54}} \times {\rm{1}}{{\rm{0}}^{ - 3}}\;mol} \right) - \left( {2.89 \times {{10}^{ - 3}}\;mol} \right)\\\ = {\rm{1}}{\rm{.65}} \times {\rm{1}}{{\rm{0}}^{ - 3}}\;mol

We can write the Avogadro number as a unit factor for N2O{N_2}O as shown below:
6.02×1023  molecules1  mol\dfrac{{6.02 \times {{10}^{23}}\;{\rm{molecules}}}}{{1\;mol}}

Finally, we can determine the number of removed molecules (x)\left( x \right) by using the removed amount and the above unit factor as follows:
1.65×10−3  mol×(6.02×1023  molecules1  mol)=9.933×1020  molecules ≈1021  molecules {\rm{1}}{\rm{.65}} \times {\rm{1}}{{\rm{0}}^{ - 3}}\;mol \times \left( {\dfrac{{6.02 \times {{10}^{23}}\;{\rm{molecules}}}}{{1\;mol}}} \right) = {\rm{9}}{\rm{.933}} \times {10^{20}}\;{\rm{molecules}}\\\ \approx {10^{21}}\;{\rm{molecules}}

Hence, the value of xx is found to be 1021  molecules{10^{21}}\;{\rm{molecules}} which makes

option D to be the correct one.

Note:
We have to be careful with the units and the exponential powers while performing the calculations.