Question

Acetic acid forms $100$% dimer in benzene by hydrogen bonding. If analytical molarity is $1$M, then its equilibrium molarity is $y\times {10}^{-1}M$. Here, $y$ is :

Answer 1

In benzene solution, the following equilibrium exists $2C{H}_{3}COOH\left(monomer\right)\leftrightharpoons \left(C{H}_{3}COOH{)}_2\right(dimer)$.
In benzene, acetic acid shows $100$% dimerization.
The molecular weight of acetic acid in benzene is $2\times 60=120g/mol$.
The equilibrium concentration of the acetic acid monomer is $0.0$ M and the equilibrium concentration of acetic acid dimer is $\frac{1}{2}=0.5M$
This is because, if only monomer of acetic acid was present, then $1$L of solution should have contained $1$ mole.
But, if dimer of acetic acid is present, then $1$ L of the solution will contain $\frac{60g{L}^{-1}}{120gmo{l}^{-1}}=0.5mol{L}^{-1}$ $=5\times {10}^{-1}M$
This is the equilibrium molarity.