' W ' g of a non-volatile electrolyte solid solute of molar mass ' M ' $\mathrm{g} \mathrm{mol}^{-1}$ when dissolved in 100 mL water, decreases vapour pressure of water from 640 mm Hg to 600 mm Hg . If aqueous solution of the electrolyte boils at 375 K and $\mathrm{K}_{\mathrm{b}}$ for water is $0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, then the mole fraction of the electrolyte solute $\left(x_2\right)$ in the solution can be expressed as

(Given : density of water $=1 \mathrm{~g} / \mathrm{mL}$ and boiling point of water $=373 \mathrm{~K}$ )

The hydrogen spectrum consists of several spectral lines in Lyman series $\left(L_1, L_2\right.$, $\mathrm{L}_3 \ldots ; \mathrm{L}_1$ has lowest energy among Lyman series). Similarly it consists of several spectral lines in Balmer series $\left(\mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3 \ldots ; \mathrm{B}_1\right.$ has lowest energy among Balmer lines). The energy of $L_1$ is $x$ times the energy of $B_1$. The value of $x$ is $\_\_\_\_$ $\times 10^{-1}$

. (Nearest integer)

Electricity is passed through an acidic solution of $\mathrm{Cu}^{2+}$ till all the $\mathrm{Cu}^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL . The total volume of oxygen evolved at STP during the entire process is $\_\_\_\_$ mL . (Nearest integer)

[Given:

$$ \begin{aligned} & \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{~s}) \mathrm{E}_{\mathrm{red}}^{\mathrm{o}}=+0.34 \mathrm{~V} \\ & \mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O} \mathrm{E}_{\mathrm{red}}^{\mathrm{o}}=+1.23 \mathrm{~V} \end{aligned} $$

Molar mass of $\mathrm{Cu}=63.54 \mathrm{~g} \mathrm{~mol}^{-1}$

Molar mass of $\mathrm{O}_2=32 \mathrm{~g} \mathrm{~mol}^{-1}$

Faraday Constant $=96500 \mathrm{C} \mathrm{mol}^{-1}$

Molar volume at $\mathrm{STP}=22.4 \mathrm{~L}$ ]

Consider two Group IV metal ions $\mathrm{X}^{2+}$ and $\mathrm{Y}^{2+}$.

A solution containing $0.01 \mathrm{M} \mathrm{X}^{2+}$ and $0.01 \mathrm{M} \mathrm{Y}^{2+}$ is saturated with $\mathrm{H}_2 \mathrm{~S}$. The pH at which the metal sulphide YS will form as a precipitate is $\_\_\_\_$ . (Nearest integer)

(Given: $\mathrm{K}_{\mathrm{sp}}(\mathrm{XS})=1 \times 10^{-22}$ at $25^{\circ} \mathrm{C}, \mathrm{K}_{\mathrm{sp}}(\mathrm{YS})=4 \times 10^{-16}$ at $25^{\circ} \mathrm{C}$, $\left[\mathrm{H}_2 \mathrm{~S}\right]=0.1 \mathrm{M}$ in solution, $\mathrm{K}_{a 1} \times \mathrm{K}_{a 2}\left(\mathrm{H}_2 \mathrm{~S}\right)=1.0 \times 10^{-21}, \log 2=0.30$, $\log 3=0.48, \log 5=0.70)$