5. Given that $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are non-negative real numbers, and
$$
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=100 \text {, }
$$
$M$ is the maximum value of $x_{1}+x_{2}, x_{2}+x_{3}, x_{3}+x_{4}, x_{4}+x_{5}$. Then the minimum value $m$ of $M$ satisfies ( ).
(A) $20<m<30$
(B) $30<m<40$
(C) $40<m<45$
(D) $45<m<50$
5.B.

According to the problem, we have
$$
M \geqslant x_{1}+x_{2}, M \geqslant x_{2}+x_{3}, M \geqslant x_{4}+x_{5} .
$$

Adding the three inequalities, we get $3 M \geqslant 100+x_{2} \geqslant 100$, which means $M \geqslant \frac{100}{3}$. When $x_{2}=x_{4}=0, x_{1}=x_{3}=x_{5}=\frac{100}{3}$, we have $M=\frac{100}{3}$. Therefore, the minimum value of $M$ is $m=\frac{100}{3}$.