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Question: \( COC{l_2} \) gas dissociates according to the equation \( COC{l_2}(g) \rightleftarrows CO(g) + C{l...

COCl2COC{l_2} gas dissociates according to the equation COCl2(g)⇄CO(g)+Cl2(g)COC{l_2}(g) \rightleftarrows CO(g) + C{l_2}(g) . When heated at 700  K700\;K the density of the gas mixture at 1.16  atm1.16\;atm and at equilibrium is 1.16g  litre−11.16g\;litre^{-1} . The degree of dissociation of COCl2COC{l_2} at 700  K700\;K is
(A) 0.28  0.28\;
(B) 0.50  0.50\;
(C) 0.72  0.72\;
(D) 0.42  0.42\;

Explanation

Solution

Hint : The formula we will be using to solve the above question is PV=nRTPV = nRT . Then we will express it in terms of density. This will help us to find the degree of dissociation with the help of the other values given in the equation.

Complete Step By Step Answer:
We all are familiar with the ideal gas law equation. It states that the product of the pressure of a gas times the volume of a gas is a constant for a given sample of gas. It can be written as:-
PV=nRTPV = nRT
where,P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant and T stands for temperature.
We are also familiar with the term Density which is mass per unit volume. The formula is:
d=MVd = \dfrac{M}{V}
The above equation can also be written as
V=MdV = \dfrac{M}{d}
Putting this above value of volume in the ideal gas law equation
PMd=nRTP\dfrac{M}{d} = nRT
PM=dRTPM = dRT
Here we have assumed the number of moles to be 11 .
Now using this derived formula we will solve the given question
M=dRTPM = \dfrac{{dRT}}{P}
P=1.16P=1.16
T=700T= 700
d=1.16d= 1.16
R=0.0821R= 0.0821
Putting the above value in the equation
M=1.16×0.0821×7001.16=57.47M = \dfrac{{1.16 \times 0.0821 \times 700}}{{1.16}}=57.47
This is the observed mass denoted by MO{M_O}
For the theoretical mass
MT=COCl2=12+16+(35.5×2)=99{M_T} = COC{l_2} = 12 + 16 + (35.5 \times 2)= 99
For degree of dissociation
α=MT−MO(n−1)MO\alpha = \dfrac{{{M_T} - {M_O}}}{{(n - 1){M_O}}}
= 99−57.4757.47=0.72\dfrac{{99 - 57.47}}{{57.47}} = 0.72
Therefore our answer is option C.   0.72\;0.72

Note :
The ideal gas equation is used only for the ideal gas and cannot be applied to the real gases. The reason for this is due to certain assumptions which are taken into consideration for the ideal gas and cannot be true for the real gas, for example volume of the real gas cannot be negligible. So the answer is only a close value of the degree of dissociation.