
The vapour density of a mixture containing \[N{O_2}\] and \[{N_2}{O_4}\] is \[38.3\] at \[27^\circ C\]. Calculate the moles of \[N{O_2}\] in \[100g\]of the mixture.
Answer
573.3k+ views
Hint: Vapour density is equal to the density of a vapour compared to that of hydrogen. It is rather calculated as the ratio of mass of a certain volume of a substance divided by mass of the same volume of hydrogen.
Complete step by step answer: The mathematical equation of vapour density can be expressed as:
\[{V_d} = \dfrac{{{M_n}(g)}}{{{M_n}({H_2})}}\]
where \[{V_d}\] = vapour density,
\[{M_n}(g)\] = mass of n number of molecules of desired gas,
\[{M_n}({H_2})\] = mass of n number of molecules of hydrogen.
The molar mass of the mixture = $2 \times {V_d} = 2 \times 38.3 = 76.6.$
For \[N{O_2}\] the molecular weight = molecular weight of \[N\] + \[2\] x molecular weight of \[O\]
= $14 + 2 \times 16 = 46$ .
For \[{N_2}{O_4}\] the molecular weight = \[2\] x molecular weight of \[N\] + \[4\] x molecular weight of \[O\]
= $2 \times 14 + 4 \times 16 = 92$ .
Let the total moles of mixture = \[1\]
Let the moles of \[N{O_2}\] = \[x\]
The moles of \[{N_2}{O_4}\] = \[1 - x\]
Moles of \[N{O_2}\] + moles of \[{N_2}{O_4}\] = moles of mixture
$46x + 92(1 - x) = 76.6$
$46x = 15.4$
$x = \dfrac{{15.4}}{{46}} = 0.335$
The percentage of \[N{O_2}\] = 33.5
And the percentage of \[{N_2}{O_4}\] = 66.5
Thus the number of moles of \[N{O_2}\] in \[100g\]of the mixture = $0.335 \times \dfrac{{100}}{{76.6}}$
= $0.437$
Note: Density and vapour density should not be confused as density is the ratio of mass of a substance and the volume of the substance while vapour density is the ratio of weight obtained by the dividing volume of gas or vapour compared to the weight of an equal volume of hydrogen. The number of moles can be calculated by dividing the mass of substance and the substance's atomic or molecular weight. Similarly the mole fraction is determined by dividing the moles of one substance in a mixture by the total number of moles of all substances in the mixture.
Complete step by step answer: The mathematical equation of vapour density can be expressed as:
\[{V_d} = \dfrac{{{M_n}(g)}}{{{M_n}({H_2})}}\]
where \[{V_d}\] = vapour density,
\[{M_n}(g)\] = mass of n number of molecules of desired gas,
\[{M_n}({H_2})\] = mass of n number of molecules of hydrogen.
The molar mass of the mixture = $2 \times {V_d} = 2 \times 38.3 = 76.6.$
For \[N{O_2}\] the molecular weight = molecular weight of \[N\] + \[2\] x molecular weight of \[O\]
= $14 + 2 \times 16 = 46$ .
For \[{N_2}{O_4}\] the molecular weight = \[2\] x molecular weight of \[N\] + \[4\] x molecular weight of \[O\]
= $2 \times 14 + 4 \times 16 = 92$ .
Let the total moles of mixture = \[1\]
Let the moles of \[N{O_2}\] = \[x\]
The moles of \[{N_2}{O_4}\] = \[1 - x\]
Moles of \[N{O_2}\] + moles of \[{N_2}{O_4}\] = moles of mixture
$46x + 92(1 - x) = 76.6$
$46x = 15.4$
$x = \dfrac{{15.4}}{{46}} = 0.335$
The percentage of \[N{O_2}\] = 33.5
And the percentage of \[{N_2}{O_4}\] = 66.5
Thus the number of moles of \[N{O_2}\] in \[100g\]of the mixture = $0.335 \times \dfrac{{100}}{{76.6}}$
= $0.437$
Note: Density and vapour density should not be confused as density is the ratio of mass of a substance and the volume of the substance while vapour density is the ratio of weight obtained by the dividing volume of gas or vapour compared to the weight of an equal volume of hydrogen. The number of moles can be calculated by dividing the mass of substance and the substance's atomic or molecular weight. Similarly the mole fraction is determined by dividing the moles of one substance in a mixture by the total number of moles of all substances in the mixture.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

