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A compound exists in the gaseous phase both as monomer (A) and dimer $(A_2)$. The molecular weight of A is 48 . In an experiment 96g of the compound was confined in a vessel of volume 33.6 litre and heated to $273^{\large\circ}C$. Calculate the pressure developed if the compound exists as dimer to the extent of $50\%$ by weight under these conditions.

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

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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)
answered Feb 15, 2014 by sharmaaparna1
 

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