The quotient space of a normed space $(X, \|\cdot\|_X)$ by a closed subspace $M\subset X$ is the normed space $(X/M, \|\cdot\|_{X/M})$ defined as follows \begin{align*} X/M= \{\dot{x}=x+M: x\in X\}\quad \text{and}\quad \|\dot{x}\|_{X/M}=\inf\big\{ \|x-y\|_{X}: y\in M\big\}=dist(x,M). \end{align*} Recall that each $\dot{x}= x+M=\{x+y: y\in M\}$ represents a equivalence class associated with the equivalence relation $\sim$ for which $x\sim y$ if and only if $x-y\in M$.
The orthogonal of $M$ is defined as follows $$M^\perp= \{\ell\in X^*: \ell(x)=0 \quad \text{for all} \quad x\in M\}= \{\ell\in X^*: M\subset \ker\ell\}. $$
Where $X^*$ is the dual of $X$.
Question Prove or disprove that the spaces $M^\perp$ and $X/M$ are isomorphic. We assume that $M$ is proper subspace of $X$.
For the case $\dim X<\infty$ it is obvious that $M^\perp$ and $X/M$ are isomorphic since $$\dim M^\perp= \dim X/M= \dim X-\dim M$$
It is well-known that $(X/M)^* \cong M^\perp$, so a space $X$ and a subspace $M$ such that $(X/M)^* \not\cong X/M$ will do the job.
Take $M=0$. Then you are asked to find a space $X$ with $X^* \not\cong X$, and there are plenty of such spaces.