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Give ${K_c}$ for is 0.04 at $25^\circ C$ .The number of moles of $PC{l_5}$ required to a 3.0 litre flask to obtain $C{l_2}$ of concentration 0.15 $M$ is
A. 1.5 mole
B. 2.1 mole
C. 6.0 mole
D. 0.45 mole

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Last updated date: 03rd Mar 2024
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IVSAT 2024
Answer
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Hint: $PC{l_5}$ decomposes to $PC{l_{{3_{}}}}$ and chlorine gas and all these remain in a state of equilibrium. Initially, $PC{l_5}$ decomposes to form the products and once the equilibrium is achieved the reaction takes place in both the forward and backward direction simultaneously. Here ${K_c}$ is called the equilibrium constant and is used to quantify the equilibrium. Here the concentration of one of the products is given so we can get to know about the concentration of the rest of the components with its help.

Complete step by step answer:
$PC{l_5}$ decomposes to $PC{l_3}$ and chlorine gas and all these remain in a state of equilibrium.
The equilibrium constant of the reaction is given as, ${K_c}$
 for $PC{l_5}(g) \rightleftharpoons PC{l_3}(g) + C{l_2}(g)$ is 0.04 at ${25^\circ }C$
Let the initial concentration of being $a$ and the amount reacted at equilibrium be $x$.
So the reaction and the initial concentration can be represented as
$PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$
$a(R),0(P),0(P)$
Let at equilibrium the amount reacted be $x$ . So,
$PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$
$a - x(R),x(P),x(P)$
The concentration of chlorine gas is given in the question as 0.15 $M$ So using the molarity relation we can find the number of moles of chlorine gas at equilibrium
$molarity(M) = \dfrac{{moles\; of \;solute}}{{Volume\; Of \;Solution(L)}}$
$ \Rightarrow 0.15 = \dfrac{{moles\;of\;C{l_2}}}{3}$
$ \Rightarrow \;Moles\;of\;C{l_2} = 0.45$
Therefore from the previous point, we can infer that the value of $x$ is 0.45
So, the equilibrium concentration becomes
$PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$
$a - 0.45(PC{l_5}),0.45(PC{l_3}),0.45(C{l_2})$
Since The equilibrium constant of the reaction is given as 0.04
We can write the equilibrium expression as
${K_c} = \dfrac{{[PC{l_3}][C{l_2}]}}{{[PC{l_5}]}}$
$ \Rightarrow 0.04 = \dfrac{{0.15 \times 0.15}}{{\left( {\dfrac{{a - 0.45}}{3}} \right)}}$
(Dividing the number of equilibrium moles with the volume of the solution at equilibrium to get the molar concentration.)
Cross multiplying and simplifying we get,
$a = 2.1$

So, the correct answer is Option B.

Note: There are other representations of the concentration of solution than molarity.
While molality is represented by $m$, Molarity is represented by $M$.
Molarity of the solution can be represented in terms of the volume of the solution rather than the mass of the solvent as In case of molality. It can be represented as
$Molality(m) = \dfrac{{{{Moles\; of \;solute}}}}{{{{Mass\; of \;solvent \;in\; kg}}}}$
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