AP EAMCET 2013 Chemistry Question Paper with Answer and Solution

(A) The height of a liquid column in a capillary tube is given by the formula: $h = \frac{2T \cos \theta}{rdg}$.
For water: $h_1 = \frac{2T_1 \cos \theta_1}{r d_1 g}$. Since $\theta_1 = 0^{\circ}$,$\cos 0^{\circ} = 1$,so $h_1 = \frac{2T_1}{r d_1 g}$.
The mass of the liquid is $m = \pi r^2 h d$. Substituting $h$,we get $m = \pi r^2 d \left( \frac{2T \cos \theta}{rdg} \right) = \frac{2 \pi r T \cos \theta}{g}$.
For the first case (water): $m_1 = \frac{2 \pi r T_1}{g} = 5 \times 10^{-3} \ kg$.
For the second liquid: $T_2 = \sqrt{2} T_1$ and $\theta_2 = 45^{\circ}$.
$m_2 = \frac{2 \pi r T_2 \cos \theta_2}{g} = \frac{2 \pi r (\sqrt{2} T_1) \cos 45^{\circ}}{g}$.
Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have $m_2 = \frac{2 \pi r (\sqrt{2} T_1) \times \frac{1}{\sqrt{2}}}{g} = \frac{2 \pi r T_1}{g}$.
Comparing $m_1$ and $m_2$,we see that $m_2 = m_1 = 5 \times 10^{-3} \ kg$.

(C) The terminal velocity $v$ of a spherical drop is given by $v = \frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}$.
Since the buoyant force is ignored,$\sigma \approx 0$,so $v \propto r^2$.
When two drops of radius $r$ combine to form a bigger drop of radius $R$,the volume is conserved:
$\frac{4}{3} \pi R^3 = 2 \times \frac{4}{3} \pi r^3$
$R^3 = 2r^3 \Rightarrow R = 2^{1/3} r$.
Let $v'$ be the terminal velocity of the bigger drop.
$\frac{v'}{v} = \frac{R^2}{r^2} = \frac{(2^{1/3} r)^2}{r^2} = 2^{2/3} = \sqrt[3]{4}$.
Therefore,$v' = \sqrt[3]{4} v$.

(A) Diborane $(B_2H_6)$ reacts with $HCl$ in the presence of $AlCl_3$ as a catalyst to undergo hydrochlorination,which results in the liberation of hydrogen gas $(H_2)$.
The chemical reaction is as follows:
$B_2H_6 + HCl \xrightarrow{AlCl_3} B_2H_5Cl + H_2 \uparrow$

(C) In three-dimensional silicates,all four oxygen atoms (corners) of each $SiO_4^{4-}$ tetrahedron are shared with other tetrahedra.
This results in a three-dimensional network structure.
Examples of such structures include various forms of silica like quartz,tridymite,and cristobalite.

(D) The reaction of sodium thiosulphate $(Na_2S_2O_3)$ with moist chlorine $(Cl_2)$ is an oxidation reaction.
In the presence of water,$Na_2S_2O_3$ is oxidized to sodium sulphate $(Na_2SO_4)$ and hydrochloric acid $(HCl)$,while sulphur $(S)$ is precipitated as a byproduct.
The balanced chemical equation is:
$Na_2S_2O_3 + Cl_2 + H_2O \longrightarrow Na_2SO_4 + 2HCl + S$
Here,$X$ corresponds to sulphur $(S)$.

(A) disproportionation reaction is a special type of redox reaction in which the same element in a given oxidation state is simultaneously oxidized and reduced.
In the reaction $3Cl_{2\text{(g)}} + 6OH^-{_{\text{(aq)}}} \rightarrow ClO_3^-{_{\text{(aq)}}} + 5Cl^-{_{\text{(aq)}}} + 3H_2O_{\text{(l)}}$,the oxidation state of chlorine changes from $0$ in $Cl_2$ to $-1$ in $Cl^-$ (reduction) and to $+5$ in $ClO_3^-$ (oxidation).
Thus,it is a disproportionation reaction.

(B) Given: Concentration $C = 0.10 \ M$,degree of ionization $\alpha = 4.0 \% = 0.04$.
For the dissociation of lactic acid $(CH_3CH(OH)COOH)$:
$CH_3CH(OH)COOH \rightleftharpoons CH_3CH(OH)COO^- + H^+$
At equilibrium,the concentrations are:
$[CH_3CH(OH)COOH] = C(1 - \alpha)$
$[CH_3CH(OH)COO^-] = C\alpha$
$[H^+] = C\alpha$
The dissociation constant $K_a$ is given by:
$K_a = \frac{[CH_3CH(OH)COO^-][H^+]}{[CH_3CH(OH)COOH]} = \frac{C\alpha \cdot C\alpha}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha}$
Substituting the values:
$K_a = \frac{0.1 \times (0.04)^2}{1 - 0.04} = \frac{0.1 \times 0.0016}{0.96} = \frac{0.00016}{0.96} = \frac{1.6 \times 10^{-4}}{0.96} \approx 1.66 \times 10^{-4}$

(D) The kinetic energy $(KE)$ of $n$ moles of an ideal gas is given by $KE = n \times \frac{3}{2} RT$.
For $4$ $g$ of $H_2$,the number of moles $n_{H_2} = \frac{4 \ g}{2 \ g/mol} = 2$ $mol$.
So,$KE_{H_2} = 2 \times \frac{3}{2} RT = 3RT$.
For $8$ $g$ of $O_2$,the number of moles $n_{O_2} = \frac{8 \ g}{32 \ g/mol} = 0.25$ $mol$ or $\frac{1}{4}$ $mol$.
So,$KE_{O_2} = \frac{1}{4} \times \frac{3}{2} RT = \frac{3}{8} RT$.
The ratio $KE_{H_2} : KE_{O_2} = 3RT : \frac{3}{8} RT = 1 : \frac{1}{8} = 8 : 1$.

(C) The quantum mechanical model of an atom is based on the dual nature of matter (wave-particle duality) proposed by $de \text{ } Broglie$.
This model considers the electron as both a particle and a wave,which is mathematically described by the $Schrodinger$ wave equation.

(B) The number of radial nodes in an orbital is given by the formula: $n - l - 1$.
For $3s$ orbital: $n = 3$ and $l = 0$. Therefore,the number of radial nodes $= 3 - 0 - 1 = 2$.
For $2p$ orbital: $n = 2$ and $l = 1$. Therefore,the number of radial nodes $= 2 - 1 - 1 = 0$.
Thus,the number of radial nodes for $3s$ and $2p$ orbitals are $2$ and $0$ respectively.

(D) Entropy $(S)$ is a measure of the randomness or disorder of a system. $A$ positive entropy change $(\Delta S > 0)$ occurs when the system becomes more disordered,such as during a phase transition from a more ordered state (liquid) to a less ordered state (gas).
In the reaction $H_2O_{(l)} \longrightarrow H_2O_{(g)}$,liquid water is converted into water vapor. Since gas molecules have much higher randomness compared to liquid molecules,the entropy of the system increases,making $\Delta S$ positive.
In other options,the reactions involve a decrease in the number of moles of gas or a transition from a disordered state to a more ordered state (like liquid to solid or gas to solid),which results in a negative entropy change.