1. Among the 95 numbers $1^{2}, 2^{2}, 3^{2}, \cdots, 95^{2}$, the numbers with an odd digit in the tens place total $\qquad$.

 untranslated part: $\qquad$
In $1^{2}, 2^{2}, \cdots, 10^{2}$, it is found through calculation that the tens digit is odd only for $4^{2}=16, 6^{2}=36$. A two-digit square number can be expressed as
$$
(10 a+b)^{2}=100 a^{2}+20 a b+b^{2},
$$

Therefore, $b$ can only be 4 or 6. That is, in every 10 consecutive numbers, there are two numbers whose tens digit is odd, hence it is deduced that in $1^{2}, 2^{2}, 3^{2}, \cdots, 95^{2}$, the number of squares with an odd tens digit is 19.