$(a)\;1\;atm\qquad(b)\;3\;atm\qquad(c)\;2\;atm\qquad(d)\;1.5\;atm$

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since A and $A_2$ are two states in gaseous phase having their weight ratio$50\%$ i.e 1:1

$\therefore Mole\;of\;A = \large\frac{96}{2}\times\large\frac{1}{48} = 1$

We have n =$ \large\frac{w}{m}$

$\therefore Mole\;of\;A_2 = \large\frac{96}{2}\times\large\frac{1}{96} = \large\frac{1}{2}$

$\therefore Total\;mole\;of\;A and A_2 are = 1+\large\frac{1}{2} = \large\frac{3}{2}$

We know that PV = nRT

$P\times33.6 = \large\frac{3}{2}\times0.0821\times546$

P = 2 atm

Hence answer is (c)

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