$x $ $\mathrm{mg}$ of pure HCl was used to make an aqueous solution. 25.0 mL of $0.1 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_2$ solution is used when the HCl solution was titrated against it. The numerical value of $x$ is $\_\_\_\_$ $\times 10^{-1}$. (Nearest integer)

Given : Molar mass of HCl and $\mathrm{Ba}(\mathrm{OH})_2$ are 36.5 and $171.0 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively.

Consider all the structural isomers with molecular formula $\mathrm{C}_5 \mathrm{H}_{11} \mathrm{Br}$ are separately treated with $\mathrm{KOH}(\mathrm{aq})$ to give respective substitution products, without any rearrangement. The number of products which can exhibit optical isomerism from these is $\_\_\_\_$。

Let the domain of the function $f(x)=\log _3 \log _5 \log _7\left(9 x-x^2-13\right)$ be the interval $(\mathrm{m}, \mathrm{n})$. Let the hyperbola $\frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$ have eccentricity $\frac{\mathrm{n}}{3}$ and the length of the latus rectum $\frac{8 \mathrm{~m}}{3}$. Then $\mathrm{b}^2-\mathrm{a}^2$ is equal to :

A rectangle is formed by the lines $x=0, y=0, x=3$ and $y=4$. Let the line L be perpendicular to $3 x+y+6=0$ and divide the area of the rectangle into two equal parts. Then the distance of the point $\left(\frac{1}{2},-5\right)$ from the line $L$ is equal to :