
Equal volume of oxygen and an unknown gas weigh \[3.00\;{\text{g}}\] and \[7.50{\text{ g}}\], respectively. Which of the following in the unknown gas?
A.\[{\text{C}}{{\text{O}}_{\text{2}}}\]
B.\[{\text{NO}}\]
C.\[{\text{N}}{{\text{O}}_2}\]
D.\[{\text{S}}{{\text{O}}_{\text{3}}}\]
Answer
137.4k+ views
Hint: For this question we must have the knowledge of Avogadro’s law. We can find unknown gas by calculating molar mass. According to Avogadro law, oxygen and the unknown gas will have the same number of molecules, from this we can calculate the molar mass of unknown gas and match it with the given options.
Formula used: \[{\text{number of moles = }}\dfrac{{{\text{mass of substance in gram}}}}{{{\text{molar mass}}}}\]
Complete step by step solution:
The above question is based on the Avogadro law, which states that “equal volume of all gases contain equal number of moles under similar condition of temperature and pressure: since pressure and temperature are not mentioned in the question so we will consider it same for both of them. In the question it has been given to us that the volume of oxygen and the unknown gas is the same. So the above condition perfectly fits the statement of Avogadro’s law. Hence the number of moles of oxygen and unknown gas will be the same. Let us calculate the number of moles of oxygen.
Mass of oxygen is \[3.00\;{\text{g}}\].
Molar mass of \[{{\text{O}}_{\text{2}}}\] is \[{\text{16 + 16 = 32 gmo}}{{\text{l}}^{ - 1}}\]
\[{{\text{n}}_{{{\text{O}}_{\text{2}}}}}{\text{ = }}\dfrac{{{\text{3 g}}}}{{{\text{32 gmo}}{{\text{l}}^1}}} = {\text{0}}{\text{.09375 moles}}\]
Let \[n\] be the number of moles of unknown gas. By statement we have \[{\text{n = }}{{\text{n}}_{{{\text{O}}_{\text{2}}}}}\].
So the number of moles of unknown gas will be \[0.09375\] and the mass is given as \[7.50{\text{ g}}\]. Putting the values in the formula:
\[{\text{0}}{\text{.09375 moles = }}\dfrac{{{\text{7}}{\text{.50 g}}}}{{{\text{molar mass in gmo}}{{\text{l}}^1}}}\]
Rearranging the above equation we will get:
\[{\text{molar mass (in gmo}}{{\text{l}}^1}{\text{) = }}\dfrac{{{\text{7}}{\text{.50 g}}}}{{0.09375{\text{ mole}}}} = 80{\text{ gmo}}{{\text{l}}^1}\]
The molar mass of unknown gas is \[ = 80{\text{ gmo}}{{\text{l}}^1}\]. Now let us check the molar mass of each of the gases given in the option.
\[{\text{Molar mass of C}}{{\text{O}}_{\text{2}}}{\text{ = 12 + 2}} \times {\text{16 = 44 gmo}}{{\text{l}}^{ - 1}}\]
\[{\text{Molar mass of NO = 14 + 16 = 30 gmo}}{{\text{l}}^{ - 1}}\]
\[{\text{Molar mass of N}}{{\text{O}}_2}{\text{ = 12 + 2}} \times {\text{16 = 46 gmo}}{{\text{l}}^{ - 1}}\]
\[{\text{Molar mass of S}}{{\text{O}}_{\text{3}}}{\text{ = 32 + 3}} \times {\text{16 = 80 gmo}}{{\text{l}}^{ - 1}}\]
Hence, option D is correct.
Note: Instead of number of moles, the examiner can even ask to calculate the number of molecules. In that case the procedure will be exactly the same. We just need to multiply the number of moles with Avogadro’s number which is constant value i.e. \[6.022 \times {10^{23}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}\]. If the number of moles of two gases will be the same then obviously the number of molecules will be the same.
Formula used: \[{\text{number of moles = }}\dfrac{{{\text{mass of substance in gram}}}}{{{\text{molar mass}}}}\]
Complete step by step solution:
The above question is based on the Avogadro law, which states that “equal volume of all gases contain equal number of moles under similar condition of temperature and pressure: since pressure and temperature are not mentioned in the question so we will consider it same for both of them. In the question it has been given to us that the volume of oxygen and the unknown gas is the same. So the above condition perfectly fits the statement of Avogadro’s law. Hence the number of moles of oxygen and unknown gas will be the same. Let us calculate the number of moles of oxygen.
Mass of oxygen is \[3.00\;{\text{g}}\].
Molar mass of \[{{\text{O}}_{\text{2}}}\] is \[{\text{16 + 16 = 32 gmo}}{{\text{l}}^{ - 1}}\]
\[{{\text{n}}_{{{\text{O}}_{\text{2}}}}}{\text{ = }}\dfrac{{{\text{3 g}}}}{{{\text{32 gmo}}{{\text{l}}^1}}} = {\text{0}}{\text{.09375 moles}}\]
Let \[n\] be the number of moles of unknown gas. By statement we have \[{\text{n = }}{{\text{n}}_{{{\text{O}}_{\text{2}}}}}\].
So the number of moles of unknown gas will be \[0.09375\] and the mass is given as \[7.50{\text{ g}}\]. Putting the values in the formula:
\[{\text{0}}{\text{.09375 moles = }}\dfrac{{{\text{7}}{\text{.50 g}}}}{{{\text{molar mass in gmo}}{{\text{l}}^1}}}\]
Rearranging the above equation we will get:
\[{\text{molar mass (in gmo}}{{\text{l}}^1}{\text{) = }}\dfrac{{{\text{7}}{\text{.50 g}}}}{{0.09375{\text{ mole}}}} = 80{\text{ gmo}}{{\text{l}}^1}\]
The molar mass of unknown gas is \[ = 80{\text{ gmo}}{{\text{l}}^1}\]. Now let us check the molar mass of each of the gases given in the option.
\[{\text{Molar mass of C}}{{\text{O}}_{\text{2}}}{\text{ = 12 + 2}} \times {\text{16 = 44 gmo}}{{\text{l}}^{ - 1}}\]
\[{\text{Molar mass of NO = 14 + 16 = 30 gmo}}{{\text{l}}^{ - 1}}\]
\[{\text{Molar mass of N}}{{\text{O}}_2}{\text{ = 12 + 2}} \times {\text{16 = 46 gmo}}{{\text{l}}^{ - 1}}\]
\[{\text{Molar mass of S}}{{\text{O}}_{\text{3}}}{\text{ = 32 + 3}} \times {\text{16 = 80 gmo}}{{\text{l}}^{ - 1}}\]
Hence, option D is correct.
Note: Instead of number of moles, the examiner can even ask to calculate the number of molecules. In that case the procedure will be exactly the same. We just need to multiply the number of moles with Avogadro’s number which is constant value i.e. \[6.022 \times {10^{23}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}\]. If the number of moles of two gases will be the same then obviously the number of molecules will be the same.
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