Let $M$ be an module over the commutative ring $R$. I'd like to ask do we have the following isomorphism?

$$Hom_R(\wedge^n_RM,R)\simeq \wedge^n_R Hom_R(M,R)$$

We can obviously see it's true for the case $M=R^m$ is a free $R$-module. But I don't know in general.