
A 100 feet long classroom maintains a seating row after every 10 feet and has doors on both front and back sides. A laughing gas $N_2O$ cylinder and a tear gas (methane) cylinder were opened simultaneously at the front and the back door respectively. Assuming both the gases were present at the same temperature and pressure and the cylinder has similar valve dimensions. If the student of nth row, from the front, simultaneously weeps and laughs, the value of n is.
A) 5
B) 4
C) 6
D) 7
Answer
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Hint:Here in this question we will use the logic of the relation between the number of rows and molecular mass of the gas. We will substitute the values in the given relation and then converge to the required answer.
Complete step by step answer:
So, both the gases will first meet between 3rd and 4th row. Therefore,
$\dfrac{{{d_{{N_2}O}}}}{{{d_{C{H_4}}}}} = \sqrt {\dfrac{{16}}{{44}}} \\$
$ = 0.6$
Therefore, while N2O travels 2.4 ft. to reach the 4th row from the front, CH4 would have travelled only 4 ft. towards the 3rd row (so, it will not reach the 3rd row by then). So, the student in the 4th row from the front will first weep and laugh simultaneously.
Therefore, the value of n=4 and the correct option is B.
Additional Information: Nitrous Oxide is also known as laughing gas. It is a chemical compound or an oxide of nitrogen with the chemical formula. It is non-flammable at room temperature with a slight metallic taste and smell. Nitrous Oxide is prepared at industrial scale by heating ammonium nitrate at about 250℃. The chemical reaction for it. The molecular mass of nitrous oxide is 44g/mol and molecular shape is linear. When we draw the Lewis structure of nitrous oxide it has a total of 16 valence electrons (5 from each of the nitrogen and 6 from oxygen).
Note:IUPAC name of Nitrous oxide is Dinitrogen monoxide. It is also commonly known as laughing gas. It is sweet-smelling, colourless and non-flammable at room temperature. Nitrous oxide is heavier than air.
Complete step by step answer:
So, both the gases will first meet between 3rd and 4th row. Therefore,
$\dfrac{{{d_{{N_2}O}}}}{{{d_{C{H_4}}}}} = \sqrt {\dfrac{{16}}{{44}}} \\$
$ = 0.6$
Therefore, while N2O travels 2.4 ft. to reach the 4th row from the front, CH4 would have travelled only 4 ft. towards the 3rd row (so, it will not reach the 3rd row by then). So, the student in the 4th row from the front will first weep and laugh simultaneously.
Therefore, the value of n=4 and the correct option is B.
Additional Information: Nitrous Oxide is also known as laughing gas. It is a chemical compound or an oxide of nitrogen with the chemical formula. It is non-flammable at room temperature with a slight metallic taste and smell. Nitrous Oxide is prepared at industrial scale by heating ammonium nitrate at about 250℃. The chemical reaction for it. The molecular mass of nitrous oxide is 44g/mol and molecular shape is linear. When we draw the Lewis structure of nitrous oxide it has a total of 16 valence electrons (5 from each of the nitrogen and 6 from oxygen).
Note:IUPAC name of Nitrous oxide is Dinitrogen monoxide. It is also commonly known as laughing gas. It is sweet-smelling, colourless and non-flammable at room temperature. Nitrous oxide is heavier than air.
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