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# When $x$ molecules are removed from $200\;mg\;{N_2}O$, $2.89 \times {10^{ - 3}}$ moles of ${N_2}O$ are left. The value of $x$ will be:A. ${10^{20}}{\rm{ molecules}}$B. ${10^{10}}{\rm{ molecules}}$C. $21 \,{\rm{ molecules}}$D. ${10^{21}}{\rm{ molecules}}$Given:- Number of removed molecules is assumed to be $x$- Initial mass of ${N_2}O$ was taken to be ${m_{{\rm{initial}}}} = 200\;mg$- Final amount of ${N_2}O$ was found to be ${n_{{\rm{final}}}} = 2.89 \times {10^{ - 3}}\;mol$

Last updated date: 20th Jun 2024
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Hint:
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, $1$ mole has been defined as the amount of matter which contains $6.02 \times {10^{23}}$ particles of matter in it. From here, we get the Avogadro number or Avogadro constant as ${N_A} = 6.02 \times {10^{23}}$. For $1$ mole of $H$ , it would be $6.02 \times {10^{23}}$ $H$ atoms, for $1$ mole of ${H_2}$ gas, it would be $6.02 \times {10^{23}}$ ${H_2}$molecules.

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

Here, we have ${N_2}O$ 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:
$\dfrac{{1\;g}}{{1000\;mg}}$

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

Now, we can write the conversion factor by using the molar mass of ${N_2}O$ which is $44.0\;gmo{l^{ - 1}}$ as follows:
$\dfrac{{1\;mol}}{{44.0\;g}}$
We will calculate the initial amount by using the initial mass and the above conversion factor as follows:
${n_{{\rm{initial}}}} = 0.2\;g \times \left( {\dfrac{{1\;mol}}{{44.0\;g}}} \right)\\ = {\rm{0}}{\rm{.00454}}\;mol\\ = {\rm{4}}{\rm{.54}} \times {\rm{1}}{{\rm{0}}^{ - 3}}\;mol$

Now, we can calculate the amount of ${N_2}O$ that was removed initial amount by subtracting final one from the initial as follows:
${n_{{\rm{removed}}}} = {n_{{\rm{initial}}}} - {n_{{\rm{final}}}}\\ = \left( {{\rm{4}}{\rm{.54}} \times {\rm{1}}{{\rm{0}}^{ - 3}}\;mol} \right) - \left( {2.89 \times {{10}^{ - 3}}\;mol} \right)\\ = {\rm{1}}{\rm{.65}} \times {\rm{1}}{{\rm{0}}^{ - 3}}\;mol$

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

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

Hence, the value of $x$ is found to be ${10^{21}}\;{\rm{molecules}}$ 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.