Question
Question: Given,1 mole mixture of \({{N}_{2}}\), \(N{{O}_{2}}\)and \({{N}_{2}}{{O}_{4}}\) has a mean molar mas...
Given,1 mole mixture of N2, NO2and N2O4 has a mean molar mass of 55.4. On heating to a temperature at which all the N2O4 dissociated as:
N2O4→2NO2
The mean molar of the mixture becomes 39.6. The molar ratio of NO2 in the original mixture is:
A. 0.4
B. 0.3
C. 0.2
D. 0.1
Solution
The mole having symbol mol is defined as the unit of measurement for the amount of substance in SI units SI stands for International System of Units. A mole of any particle or substance is defined as containing exactly 6.02214076×1023 particles which may be atoms, molecules, electrons or ions.
Complete answer:
Mole ratio is defined as the ratio between the amounts in moles of any two compounds involved in a chemical reaction. Mole ratios are used as conversion factors between products and reactants in many chemistry equations. The mole ratio can be determined in balanced reactions by examining the coefficients given in front of their formulas.
Let us consider the number of moles of N2 is given by X, number of moles of NO2 by Y and number of moles of N2O4 is given by Z.
Total moles = 1(Given)
So X+Y+Z=1(Equation 1)
Molar mass of N2= 28
Molar mass of NO2= 46
Molar mass of N2O4= 92
Mean average of the molar mass = X+Y+Z28X+46Y+92Z=55.6
As from Equation 1 we know that X+Y+Z=1
∴28X+46Y+92Z=55.6 (Equation 2)
Moles of NO2formed from N2O4= 2z
Hence after reaction we can say that NO2= Y+2Z
Now mean average of the molar mass after reaction is given by
X+Y+2Z28X+46Y+92Z = 39.6 (Equation 3)
Now by solving all the three equations we can find the values of X, Y and Z where
X = 0.5, Y = 0.1 and Z = 0.4
Molar ratio of NO2 is 0.1.
This option D is the correct answer.
Note:
Molar ratio has one main disadvantage that it can only be applied for balanced chemical reactions these are those reactions in which number of atoms for each elements in the reaction and total charge is same for both reactants and products while for unbalanced reactions molar ratio concept is of no use but this type of reactions can be balanced by hit and trial method.