500 mL of $0.2 \mathrm{M} \mathrm{MnO}_4^{-}$solution in basic medium when mixed with 500 mL of 1.5 M KI solution, oxidises iodide ions to liberate molecular iodine. This liberated iodine is then titrated with a standard $x \mathrm{M}$ thiosulphate solution in presence of starch till the end point. If 300 mL of thiosulphate was consumed, then the value of $x$ is $\_\_\_\_$。

In a closed flask at 600 K , one mole of $\mathrm{X}_2 \mathrm{Y}_4(\mathrm{~g})$ attains equilibrium as given below :

$$ \mathrm{X}_2 \mathrm{Y}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{XY}_2(\mathrm{~g}) $$

At equilibrium, $75 \% \mathrm{X}_2 \mathrm{Y}_4(\mathrm{~g})$ was dissociated and the total pressure is 1 atm . The magnitude of $\Delta_{\mathrm{r}} \mathrm{G}^{\ominus}$ (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) at this temperature is $\_\_\_\_$ . (Nearest Integer)

(Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} ; \ln 10=2.3, \log 2=0.3, \log 3=0.48, \log 5=0.69, \log 7=0.84$ )

Decomposition of a hydrocarbon follows the equation $\mathrm{k}=\left(5.5 \times 10^{11} \mathrm{~s}^{-1}\right) \mathrm{e}^{\frac{-28000 \mathrm{~K}}{\mathrm{~T}}}$. The activation energy of reaction is $\_\_\_\_$ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest Integer)

Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as $f(x)=\frac{2 x^2-3 x+2}{3 x^2+x+3}$. Then $f$ is :