AP EAMCET 2018 Chemistry Question Paper with Answer and Solution

(D) The reaction is a redox titration between $KMnO_4$ and $H_2O_2$ in acidic medium.
For $KMnO_4$,the change in oxidation state of $Mn$ is from $+7$ to $+2$,so the n-factor is $5$.
For $H_2O_2$,the change in oxidation state of $O$ is from $-1$ to $0$,so the n-factor is $2$.
At the equivalence point,the number of milliequivalents of $H_2O_2$ equals the number of milliequivalents of $KMnO_4$.
$N_1 V_1 = N_2 V_2$
Here,$N_2 = M_2 \times \text{n-factor} = 0.02 \times 5 = 0.1 \ N$.
$V_2 = 16 \ mL$,$V_1 = 20 \ mL$.
$N_1 = \frac{N_2 V_2}{V_1} = \frac{0.1 \times 16}{20} = \frac{1.6}{20} = 0.08 \ N = 8 \times 10^{-2} \ N$.

(D) The balanced chemical equation is:
$3Na_2SO_3 + K_2Cr_2O_7 + 4H_2SO_4 \rightarrow 3Na_2SO_4 + K_2SO_4 + Cr_2(SO_4)_3 + 4H_2O$
Moles of $Na_2SO_3 = 0.15 \ M \times 1 \ L = 0.15 \ mol$
Moles of $K_2Cr_2O_7 = 0.2 \ M \times 0.5 \ L = 0.1 \ mol$
From the stoichiometry,$3 \ mol$ of $Na_2SO_3$ react with $1 \ mol$ of $K_2Cr_2O_7$.
Moles of $K_2Cr_2O_7$ required for $0.15 \ mol$ of $Na_2SO_3 = \frac{0.15}{3} = 0.05 \ mol$.
Moles of $K_2Cr_2O_7$ remaining $= 0.1 \ mol - 0.05 \ mol = 0.05 \ mol$.

(B) $LiNO_3$ decomposes as: $2 LiNO_3 \xrightarrow{\Delta} Li_2O + 2 NO_2 + \frac{1}{2} O_2$
$NaNO_3$ decomposes as: $2 NaNO_3 \xrightarrow{\Delta} 2 NaNO_2 + O_2$
$Ba(NO_3)_2$ decomposes as: $2 Ba(NO_3)_2 \xrightarrow{\Delta} 2 BaO + 4 NO_2 + O_2$
$Be(NO_3)_2$ decomposes as: $2 Be(NO_3)_2 \xrightarrow{\Delta} 2 BeO + 4 NO_2 + O_2$
Thus,$NaNO_3$ forms its nitrite $(NaNO_2)$ and $O_2$ upon heating,not its oxide.

(D) As we move down group $2$ from $Mg(OH)_2$ to $Ba(OH)_2$:
$1$. The basic character increases because the metallic character and the ease of releasing $OH^-$ ions increase due to the increase in atomic size.
$2$. The chemical stability (thermal stability) increases as the electropositive character of the metal increases,making the $M-O$ bond stronger relative to the $O-H$ bond.
$3$. The solubility in water increases because the lattice energy decreases more rapidly than the hydration energy as the size of the cation increases.
Therefore,all three properties increase down the group.

(A) $1$. Thermal decomposition of calcium carbonate: $CaCO_3 \stackrel{\Delta}{\rightleftharpoons} CO_2 + CaO (X)$.
$2$. Slaking of lime: $CaO (X) + H_2O \longrightarrow Ca(OH)_2 (Y)$.
$3$. Reaction with chlorine to form bleaching powder: $Ca(OH)_2 (Y) + Cl_2 \longrightarrow CaOCl_2 (Z) + H_2O$.
Thus,$X = CaO$,$Y = Ca(OH)_2$,and $Z = CaOCl_2$.

(B) Oxoacid salts are salts derived from oxoacids (acids containing oxygen).
From the given list:
$(i)$ $CaCO_3$ is a salt of carbonic acid $(H_2CO_3)$.
$(ii)$ $MgSO_4$ is a salt of sulfuric acid $(H_2SO_4)$.
$(iv)$ $Sr(NO_3)_2$ is a salt of nitric acid $(HNO_3)$.
Compounds $(iii)$ $BaCl_2$,$(v)$ $MgBr_2$,and $(vi)$ $MgCl_2$ are halides,which are not oxoacid salts.
Therefore,the oxoacid salts are $(i)$,$(ii)$,and $(iv)$.

(A) The oxides and hydroxides of both $Be$ and $Al$ are amphoteric in nature,not basic.
Both $Be$ and $Al$ exhibit diagonal relationship due to similar ionic potential ($charge/size$ ratio).
$Be$ and $Al$ both form complexes (e.g.,$[BeF_4]^{2-}$ and $[AlF_6]^{3-}$).
Their chlorides ($BeCl_2$ and $Al_2Cl_6$) exist as bridged structures in the vapour phase and are covalent,making them soluble in organic solvents.
Therefore,the statement that they form basic oxides and hydroxides is incorrect as they are amphoteric.

(A) $Be$ and $Al$ exhibit a diagonal relationship,leading to many similar properties.
Both metals form covalent compounds and their chlorides act as Lewis acids.
Both metals dissolve in alkalies to form soluble complexes (beryllates and aluminates).
The oxides and hydroxides of both $Be$ and $Al$ are amphoteric in nature,not basic.
Therefore,the statement that both form basic oxides and hydroxides is incorrect.

(D) The inputs to the two $OR$ gates are $A$ and $B$. Therefore,the outputs of the two $OR$ gates are $Y_1 = A + B$ and $Y_2 = A + B$.
These outputs $Y_1$ and $Y_2$ are fed into a $NAND$ gate. The output $Y$ of the $NAND$ gate is given by:
$Y = \overline{Y_1 \cdot Y_2}$
Substituting the values of $Y_1$ and $Y_2$:
$Y = \overline{(A + B) \cdot (A + B)}$
Using the Boolean identity $X \cdot X = X$,we have $(A + B) \cdot (A + B) = A + B$.
Therefore,$Y = \overline{A + B}$.
The expression $\overline{A + B}$ represents the output of a $NOR$ gate.
Thus,the equivalent logic gate is a $NOR$ gate.

(B) The family of all circles in the $XY$ plane with radius $5$ units and center $(x_1, y_1)$ is given by:
$(x - x_1)^2 + (y - y_1)^2 = 25$ ... $(i)$
Differentiating with respect to $x$:
$2(x - x_1) + 2(y - y_1)y' = 0$
$(x - x_1) = -(y - y_1)y'$ ... (ii)
Differentiating again with respect to $x$:
$1 - [(y')^2 + (y - y_1)y''] = 0$
$(y - y_1) = \frac{1 - (y')^2}{y''}$ ... (iii)
Substituting $(y - y_1)$ from (iii) into (ii):
$(x - x_1) = -\left(\frac{1 - (y')^2}{y''}\right)y' = \frac{(y')^3 - y'}{y''}$
Substituting $(x - x_1)$ and $(y - y_1)$ into $(i)$:
$\left(\frac{(y')^3 - y'}{y''}\right)^2 + \left(\frac{1 - (y')^2}{y''}\right)^2 = 25$
$\frac{(y')^2((y')^2 - 1)^2 + (1 - (y')^2)^2}{(y'')^2} = 25$
$\frac{(1 - (y')^2)^2 [ (y')^2 + 1 ]}{(y'')^2} = 25$
$(1 - (y')^2)^2 (1 + (y')^2) = 25(y'')^2$
Here,the highest order derivative is $y''$,so the order $m = 2$.
The power of the highest order derivative is $2$,so the degree $l = 2$.
Given $l = 2$ and $m = 2$,we have:
$2l + 3m = 2(2) + 3(2) = 4 + 6 = 10$.

(C) Given differential equation: $\cos ^2 x \frac{d y}{d x}-(\tan 2 x) y=\cos ^4 x$.
Dividing by $\cos ^2 x$,we get: $\frac{d y}{d x} - \frac{\tan 2 x}{\cos ^2 x} y = \cos ^2 x$.
This is a linear differential equation of the form $\frac{d y}{d x} + P y = Q$,where $P = -\frac{\tan 2 x}{\cos ^2 x}$ and $Q = \cos ^2 x$.
Integrating factor $IF = e^{\int P dx} = e^{\int -\frac{\tan 2 x}{\cos ^2 x} dx}$.
Let $I = \int -\frac{\tan 2 x}{\cos ^2 x} dx = \int -\frac{\sin 2 x}{\cos 2 x \cdot \cos ^2 x} dx$.
Using $\cos^2 x = \frac{1+\cos 2 x}{2}$,we get $I = \int -\frac{2 \sin 2 x}{\cos 2 x (1+\cos 2 x)} dx$.
Let $\cos 2 x = t$,then $-2 \sin 2 x dx = dt$.
$I = \int \frac{dt}{t(t+1)} = \int (\frac{1}{t} - \frac{1}{t+1}) dt = \ln|t| - \ln|t+1| = \ln|\frac{t}{t+1}|$.
Thus,$IF = \frac{\cos 2 x}{\cos 2 x + 1} = \frac{\cos 2 x}{2 \cos^2 x}$.
The solution is $y \cdot IF = \int Q \cdot IF dx + c$.
$y \cdot \frac{\cos 2 x}{2 \cos^2 x} = \int \cos^2 x \cdot \frac{\cos 2 x}{2 \cos^2 x} dx + c = \frac{1}{2} \int \cos 2 x dx + c = \frac{1}{4} \sin 2 x + c$.
$y = \frac{2 \cos^2 x}{\cos 2 x} (\frac{\sin 2 x + 4c}{4}) = \frac{1}{2} \frac{\cos^2 x}{\cos 2 x} (\sin 2 x + C)$.
Using $\cos 2 x = \cos^2 x - \sin^2 x = \cos^2 x (1 - \tan^2 x)$,we get $y = \frac{1}{2} \frac{\sin 2 x + C}{1 - \tan^2 x}$.

(C) Let the position vectors be $\vec{a} = \hat{i}+2\hat{j}+2\hat{k}$,$\vec{b} = 2\hat{i}-\hat{j}$,$\vec{c} = \hat{i}+\hat{j}+3\hat{k}$,and $\vec{d} = 4\hat{j}+5\hat{k}$.
First,find the vectors representing the sides:
$\vec{AB} = \vec{b} - \vec{a} = (2-1)\hat{i} + (-1-2)\hat{j} + (0-2)\hat{k} = \hat{i} - 3\hat{j} - 2\hat{k}$.
$\vec{BC} = \vec{c} - \vec{b} = (1-2)\hat{i} + (1-(-1))\hat{j} + (3-0)\hat{k} = -\hat{i} + 2\hat{j} + 3\hat{k}$.
$\vec{CD} = \vec{d} - \vec{c} = (0-1)\hat{i} + (4-1)\hat{j} + (5-3)\hat{k} = -\hat{i} + 3\hat{j} + 2\hat{k}$.
$\vec{DA} = \vec{a} - \vec{d} = (1-0)\hat{i} + (2-4)\hat{j} + (2-5)\hat{k} = \hat{i} - 2\hat{j} - 3\hat{k}$.
Note that $\vec{AB} = -\vec{CD}$ and $\vec{BC} = -\vec{DA}$,which confirms $ABCD$ is a parallelogram.
Now check the lengths of the sides:
$|\vec{AB}| = \sqrt{1^2 + (-3)^2 + (-2)^2} = \sqrt{1+9+4} = \sqrt{14}$.
$|\vec{BC}| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1+4+9} = \sqrt{14}$.
Since adjacent sides are equal $(|\vec{AB}| = |\vec{BC}| = \sqrt{14})$,the parallelogram is a rhombus.

(C) Given that $a, b, c$ are unit vectors,so $|a|=|b|=|c|=1$.
We know that $|a-b|^2 = |a|^2 + |b|^2 - 2(a \cdot b) = 1 + 1 - 2(a \cdot b) = 2 - 2(a \cdot b)$.
Similarly,$|b-c|^2 = 2 - 2(b \cdot c)$ and $|c-a|^2 = 2 - 2(c \cdot a)$.
Adding these,we get:
$|a-b|^2+|b-c|^2+|c-a|^2 = 6 - 2(a \cdot b + b \cdot c + c \cdot a) \quad \dots (i)$
To maximize this expression,we need to minimize $(a \cdot b + b \cdot c + c \cdot a)$.
We know that $|a+b+c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) \ge 0$.
Since $|a|=|b|=|c|=1$,we have $1 + 1 + 1 + 2(a \cdot b + b \cdot c + c \cdot a) \ge 0$,which implies $2(a \cdot b + b \cdot c + c \cdot a) \ge -3$.
The minimum value of $2(a \cdot b + b \cdot c + c \cdot a)$ is $-3$.
Substituting this into equation $(i)$:
$k = 6 - (-3) = 9$.
Now,we need to calculate $k(2|a|^2+3|b|^2-4|c|^2)$.
Since $|a|=|b|=|c|=1$,we have $|a|^2=|b|^2=|c|^2=1$.
Thus,$k(2(1)+3(1)-4(1)) = 9(2+3-4) = 9(1) = 9$.

(D) Let the angles made by the line with the $X$,$Y$,and $Z$-axes be $\alpha$,$\beta$,and $\gamma$ respectively.
Given:
$\alpha = \tan ^{-1} \sqrt{7} \implies \tan \alpha = \sqrt{7}$.
Since $\sec^2 \alpha = 1 + \tan^2 \alpha = 1 + 7 = 8$,we have $\cos^2 \alpha = \frac{1}{8}$.
$\beta = \tan ^{-1} \frac{\sqrt{5}}{\sqrt{3}} \implies \tan \beta = \frac{\sqrt{5}}{\sqrt{3}}$.
Since $\sec^2 \beta = 1 + \tan^2 \beta = 1 + \frac{5}{3} = \frac{8}{3}$,we have $\cos^2 \beta = \frac{3}{8}$.
We know that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.
Substituting the values:
$\frac{1}{8} + \frac{3}{8} + \cos^2 \gamma = 1$
$\frac{4}{8} + \cos^2 \gamma = 1$
$\frac{1}{2} + \cos^2 \gamma = 1$
$\cos^2 \gamma = \frac{1}{2}$
$\cos \gamma = \pm \frac{1}{\sqrt{2}}$
Thus,$\gamma = \frac{\pi}{4}$ or $\gamma = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

(D) The given equation of the plane is $x + 3y + 2z = 9$.
Dividing by $9$,we get $\frac{x}{9} + \frac{y}{3} + \frac{z}{9/2} = 1$.
Thus,the intercepts made by the plane with the coordinate axes are $a = 9$,$b = 3$,and $c = 9/2$.
The direction ratios of the normal to the required plane are $(9, 3, 9/2)$.
Multiplying by $2$,we can take the direction ratios as $(18, 6, 9)$,or simply $(6, 2, 3)$.
The plane passes through the point $(3, 5, 7)$.
The equation of the plane is $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$.
Substituting the values: $6(x - 3) + 2(y - 5) + 3(z - 7) = 0$.
$6x - 18 + 2y - 10 + 3z - 21 = 0$.
$6x + 2y + 3z - 49 = 0$.
Therefore,the equation of the plane is $6x + 2y + 3z = 49$.

(B) Total number of $4$-digit numbers formed using digits ${0, 1, 2, 3, 4, 6}$ without repetition:
Thousands place can be filled in $5$ ways (excluding $0$).
Hundreds place can be filled in $5$ ways.
Tens place can be filled in $4$ ways.
Units place can be filled in $3$ ways.
Total $= 5 \times 5 \times 4 \times 3 = 300$.
$A$ number is divisible by $4$ if the number formed by its last two digits is divisible by $4$. The possible last two-digit combinations from ${0, 1, 2, 3, 4, 6}$ are:
$04, 12, 16, 20, 24, 32, 36, 40, 60, 64$.
Case $1$: Last two digits are $04, 20, 24, 40, 60, 64$ (contain $0$ or $4$).
- If the last two digits are $04, 20, 40, 60$:
Remaining $2$ places can be filled in $4 \times 3 = 12$ ways each.
Total $= 4 \times 12 = 48$ ways.
- If the last two digits are $24, 64$:
Thousands place cannot be $0$ or the digits used.
Thousands place has $3$ choices,Hundreds place has $3$ choices.
Total $= 2 \times (3 \times 3) = 18$ ways.
Case $2$: Last two digits are $12, 16, 32, 36$:
Thousands place cannot be $0$ or the digits used.
Thousands place has $3$ choices,Hundreds place has $3$ choices.
Total $= 4 \times (3 \times 3) = 36$ ways.
Total favorable outcomes $= 48 + 18 + 36 = 102$.
Probability $= \frac{102}{300} = \frac{17}{50}$.

(A) Total number of cards = $52$.
The total number of ways to draw $4$ cards from $52$ is $^{52}C_4 = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270725$.
For the favourable case,we need exactly two cards from one suit and one card each from two other different suits.
Step $1$: Select $1$ suit out of $4$ to get $2$ cards: $^4C_1 = 4$ ways.
Step $2$: Select $2$ cards from the $13$ cards of that suit: $^{13}C_2 = \frac{13 \times 12}{2} = 78$ ways.
Step $3$: Select $2$ other suits out of the remaining $3$ suits: $^3C_2 = 3$ ways.
Step $4$: Select $1$ card each from these $2$ selected suits: $^{13}C_1 \times ^{13}C_1 = 13 \times 13 = 169$ ways.
Total favourable ways = $4 \times 78 \times 3 \times 169 = 157872$.
Probability = $\frac{157872}{270725} = \frac{72 \times 169}{425 \times 49}$.

(C) Given $n=5$ and $p=\frac{3}{4}$,so $q=1-p=\frac{1}{4}$.
$\alpha = \frac{1}{9} P(X \geq 3) = \frac{1}{9} [P(X=3) + P(X=4) + P(X=5)]$
$= \frac{1}{9} [^5C_3(\frac{3}{4})^3(\frac{1}{4})^2 + ^5C_4(\frac{3}{4})^4(\frac{1}{4})^1 + ^5C_5(\frac{3}{4})^5]$
$= \frac{1}{9} [10 \cdot \frac{27}{1024} + 5 \cdot \frac{81}{1024} + 1 \cdot \frac{243}{1024}] = \frac{1}{9} [\frac{270+405+243}{1024}] = \frac{1}{9} [\frac{918}{1024}] = \frac{102}{1024}$.
$\beta = P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$
$= ^5C_0(\frac{3}{4})^0(\frac{1}{4})^5 + ^5C_1(\frac{3}{4})^1(\frac{1}{4})^4 + ^5C_2(\frac{3}{4})^2(\frac{1}{4})^3$
$= 1 \cdot \frac{1}{1024} + 5 \cdot \frac{3}{1024} + 10 \cdot \frac{9}{1024} = \frac{1+15+90}{1024} = \frac{106}{1024}$.
Now,$256(\beta-\alpha) = 256(\frac{106}{1024} - \frac{102}{1024}) = 256(\frac{4}{1024}) = 256(\frac{1}{256}) = 1$.

(D) We know that the sum of probabilities in a probability distribution is $1$.
$\Sigma P(X = x_i) = 1$
$\Rightarrow \frac{1}{6} + k + \frac{1}{4} + k + \frac{1}{6} = 1$
$\Rightarrow 2k + \frac{2}{6} + \frac{1}{4} = 1$
$\Rightarrow 2k + \frac{1}{3} + \frac{1}{4} = 1$
$\Rightarrow 2k + \frac{7}{12} = 1$
$\Rightarrow 2k = 1 - \frac{7}{12} = \frac{5}{12}$
$\Rightarrow k = \frac{5}{24}$
Now,the mean $E(X) = \Sigma x_i P(x_i)$:
$E(X) = (-2)(\frac{1}{6}) + (-1)(\frac{5}{24}) + (0)(\frac{1}{4}) + (1)(\frac{5}{24}) + (2)(\frac{1}{6})$
$E(X) = -\frac{2}{6} - \frac{5}{24} + 0 + \frac{5}{24} + \frac{2}{6} = 0$
Now,$E(X^2) = \Sigma x_i^2 P(x_i)$:
$E(X^2) = (-2)^2(\frac{1}{6}) + (-1)^2(\frac{5}{24}) + (0)^2(\frac{1}{4}) + (1)^2(\frac{5}{24}) + (2)^2(\frac{1}{6})$
$E(X^2) = 4(\frac{1}{6}) + 1(\frac{5}{24}) + 0 + 1(\frac{5}{24}) + 4(\frac{1}{6})$
$E(X^2) = \frac{4}{6} + \frac{5}{24} + \frac{5}{24} + \frac{4}{6} = \frac{8}{6} + \frac{10}{24} = \frac{4}{3} + \frac{5}{12} = \frac{16+5}{12} = \frac{21}{12} = \frac{7}{4}$
Variance $\text{Var}(X) = E(X^2) - [E(X)]^2$
$\text{Var}(X) = \frac{7}{4} - (0)^2 = \frac{7}{4}$

(B) Given: $p^{\circ}_{benzene} = 50 \ mm \ Hg$,$p^{\circ}_{toluene} = 40 \ mm \ Hg$.
Number of moles of benzene $(n_b) = \frac{117 \ g}{78 \ g \ mol^{-1}} = 1.5 \ mol$.
Number of moles of toluene $(n_t) = \frac{46 \ g}{92 \ g \ mol^{-1}} = 0.5 \ mol$.
Total moles $= 1.5 + 0.5 = 2.0 \ mol$.
Mole fraction of benzene $(\chi_b) = \frac{1.5}{2.0} = 0.75$.
Mole fraction of toluene $(\chi_t) = \frac{0.5}{2.0} = 0.25$.
Partial pressure of benzene $(p_b) = p^{\circ}_b \times \chi_b = 50 \times 0.75 = 37.5 \ mm \ Hg$.
Partial pressure of toluene $(p_t) = p^{\circ}_t \times \chi_t = 40 \times 0.25 = 10 \ mm \ Hg$.
Total vapour pressure $(p_T) = p_b + p_t = 37.5 + 10 = 47.5 \ mm \ Hg$.
Mole fraction of toluene in vapour phase $(y_t) = \frac{p_t}{p_T} = \frac{10}{47.5} \approx 0.21$.

(B) $\Delta T_f = K_f \times m$
$\Delta T_f = 0 - (-0.93) = 0.93 \ K$
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{36 / M}{1.2} = \frac{30}{M}$
$0.93 = 1.86 \times \frac{30}{M}$
$M = \frac{1.86 \times 30}{0.93} = 2 \times 30 = 60 \ g \ mol^{-1}$
Empirical formula mass of $CH_2O = 12 + 2(1) + 16 = 30 \ g \ mol^{-1}$
$n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{60}{30} = 2$
Molecular formula $= (CH_2O)_2 = C_2H_4O_2$

(C) Initial volume $(V_1) = 50 \ mL$
Initial normality $(N_1) = 0.1 \ N$
Final volume $(V_2) = 50 \ mL + 150 \ mL = 200 \ mL$
Using the dilution formula: $N_1 \times V_1 = N_2 \times V_2$
$0.1 \times 50 = N_2 \times 200$
$N_2 = \frac{0.1 \times 50}{200} = 0.025 \ N = \frac{1}{40} \ N$
For $Na_2CO_3$,the valence factor $(Z)$ is $2$ (total positive charge of $Na^+$ ions).
Relation between Normality and Molarity: $N = M \times Z$
$M = \frac{N}{Z} = \frac{1/40}{2} = \frac{1}{80} \ M$
Therefore,the molarity of the resultant solution is $M/80$.

(C) Given,
Molarity of $HCl$ $(M_1) = 0.1 \ M$
Volume of $Na_2CO_3$ solution $(V_2) = 20 \ mL$
Mass of $Na_2CO_3 = 0.106 \ g$
Molar mass of $Na_2CO_3 = (2 \times 23) + 12 + (3 \times 16) = 106 \ g/mol$
Moles of $Na_2CO_3 = \frac{0.106 \ g}{106 \ g/mol} = 0.001 \ mol$
Molarity of $Na_2CO_3$ $(M_2) = \frac{0.001 \ mol}{0.020 \ L} = 0.05 \ M$
The neutralization reaction is: $Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$
Using the equivalence relation: $n_{HCl} = 2 \times n_{Na_2CO_3}$
$M_1 \times V_1 = 2 \times (M_2 \times V_2)$
$0.1 \times V_1 = 2 \times (0.05 \times 20)$
$0.1 \times V_1 = 2 \times 1 = 2$
$V_1 = \frac{2}{0.1} = 20 \ mL$

(C) The pressure $P$ of a gas mixture is given by $P = \frac{nRT}{V}$. Since $R, T,$ and $V$ are constant,$P \propto n_{total}$.
In the first container,the total moles $n_1 = n_{H_2} + n_{He} + n_{O_2} = \frac{16}{2} + \frac{16}{4} + \frac{16}{32} = 8 + 4 + 0.5 = 12.5 \ mol$.
Given $P_1 = 10 \ atm$,so $10 \propto 12.5$.
In the second container,the total moles $n_2 = n_{He} + n_{O_2} = \frac{16}{4} + \frac{16}{32} = 4 + 0.5 = 4.5 \ mol$.
Therefore,the new pressure $P_2 = \frac{n_2}{n_1} \times P_1 = \frac{4.5}{12.5} \times 10 = 0.36 \times 10 = 3.6 \ atm$.

(B) Given: Amount of gas $n = 4.0 \ mol$,Volume $V = 5 \ dm^3$,Pressure $p = 3.32 \ bar$,and Gas constant $R = 0.083 \ bar \ dm^3 \ K^{-1} \ mol^{-1}$.
Using the ideal gas equation $p \ V = n \ R \ T$,we can solve for temperature $T$:
$T = \frac{p \ V}{n \ R}$
Substituting the values:
$T = \frac{3.32 \times 5}{4.0 \times 0.083}$
$T = \frac{16.6}{0.332} = 50 \ K$.
Therefore,the temperature of the gas is $50 \ K$.

(B) According to Graham's law of diffusion,the rate of diffusion of a gas is inversely proportional to the square root of its molar mass $(M_m)$.
$\frac{r_X}{r_{hydrocarbon}} = \sqrt{\frac{M_{hydrocarbon}}{M_X}}$
Given that $r_X = 3\sqrt{3} \times r_{hydrocarbon}$ and $M_{hydrocarbon} = 54 \ g \ mol^{-1}$.
Substituting the values:
$3\sqrt{3} = \sqrt{\frac{54}{M_X}}$
Squaring both sides:
$(3\sqrt{3})^2 = \frac{54}{M_X}$
$9 \times 3 = \frac{54}{M_X}$
$27 = \frac{54}{M_X}$
$M_X = \frac{54}{27} = 2 \ g \ mol^{-1}$

(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.
Therefore,$\frac{r_X}{r_{\text{gas}}} = \sqrt{\frac{M_{\text{gas}}}{M_X}}$.
Given that $r_X = 3 \sqrt{3} \times r_{\text{gas}}$,we have $\frac{r_X}{r_{\text{gas}}} = 3 \sqrt{3}$.
Substituting the values: $3 \sqrt{3} = \sqrt{\frac{54}{M_X}}$.
Squaring both sides: $(3 \sqrt{3})^2 = \frac{54}{M_X}$.
$27 = \frac{54}{M_X}$.
$M_X = \frac{54}{27} = 2 \ g \ mol^{-1}$.

(C) The most probable speed $(v_{mp})$ is given by the formula $v_{mp} = \sqrt{\frac{2RT}{M}}$.
At a constant temperature,$v_{mp} \propto \frac{1}{\sqrt{M}}$.
From the given graph,the order of most probable speeds is $r_2 < r_1 < r_3$.
Since $v_{mp}$ is inversely proportional to the square root of the molar mass,a lower speed corresponds to a higher molar mass.
Therefore,the order of molar masses is $M_2 > M_1 > M_3$.

(B) Given,$U_{rms} = 3000 \ ms^{-1}$.
The formula for $RMS$ speed is $U_{rms} = \sqrt{\frac{3RT}{M}}$.
Squaring both sides,we get $U_{rms}^2 = \frac{3RT}{M}$.
For nitrogen $(N_2)$,the molar mass $M = 28 \ g/mol = 28 \times 10^{-3} \ kg/mol$.
Substituting the values: $(3000)^2 = \frac{3RT}{28 \times 10^{-3}}$.
$9 \times 10^6 = \frac{3RT}{28 \times 10^{-3}}$.
$3RT = 9 \times 10^6 \times 28 \times 10^{-3} = 252 \times 10^3 \ J/mol$.
The kinetic energy $(K.E.)$ for $1$ mole of an ideal gas is given by $K.E. = \frac{3}{2} RT$.
$K.E. = \frac{1}{2} \times (3RT) = \frac{1}{2} \times 252 \times 10^3 \ J/mol = 126 \times 10^3 \ J/mol = 126 \ kJ$.

(D) The most probable speed $(u_{mp})$ is given by the formula: $u_{mp} = \sqrt{\frac{2RT}{M}}$.
Given $u_{mp} = 400 \ ms^{-1}$ and molar mass of methane $(M)$ = $16 \ g \ mol^{-1} = 0.016 \ kg \ mol^{-1}$.
Squaring both sides: $u_{mp}^2 = \frac{2RT}{M} \Rightarrow \frac{RT}{M} = \frac{u_{mp}^2}{2}$.
Substituting the values: $\frac{RT}{0.016} = \frac{400^2}{2} = \frac{160000}{2} = 80000$.
Thus,$RT = 80000 \times 0.016 = 1280 \ J \ mol^{-1}$.
The kinetic energy $(KE)$ of one mole of an ideal gas is given by: $KE = \frac{3}{2} RT$.
$KE = \frac{3}{2} \times 1280 = 3 \times 640 = 1920 \ J$.

(A) For element $X$: Mass number $A = 52$. Let the number of protons be $p$. Number of neutrons $n = p + 0.166p = 1.166p$. Since $A = p + n$,we have $52 = p + 1.166p = 2.166p$. Thus,$p = 52 / 2.166 \approx 24$. The element with atomic number $24$ is $Cr$.
For element $Z$: Mass number $A = 75$. Let the number of protons be $p'$. Number of neutrons $n' = p' + 0.273p' = 1.273p'$. Since $A = p' + n'$,we have $75 = p' + 1.273p' = 2.273p'$. Thus,$p' = 75 / 2.273 \approx 33$. The element with atomic number $33$ is $As$.

(D) The atomic number $(Z)$ of $Uuq$ is $114$,so the number of protons $(p)$ is $114$.
Let the number of neutrons be $n$.
According to the problem,the number of neutrons is $12.8 \%$ more than the number of protons:
$n = p + 0.128 \times p = 1.128 \times p$
$n = 1.128 \times 114 = 128.592 \approx 129$.
The mass number $(A)$ is the sum of protons and neutrons:
$A = p + n = 114 + 129 = 243$.
Therefore,the approximate mass number is $243$.

(C) For a hydrogen atom, the radius of the $n^{th}$ orbit is given by $r_n = n^2 \times a_0$, where $a_0 = 52.9 \ pm$.
Given $r_n = 476.1 \ pm$, we have $n^2 = 476.1 / 52.9 = 9$, so $n = 3$.
The energy of an electron in the $n^{th}$ orbit is given by $E_n = E_1 / n^2$.
Substituting the values, $E_3 = -2.18 \times 10^{-18} \ J / 3^2 = -2.18 \times 10^{-18} / 9 \ J$.
$E_3 = -0.2422 \times 10^{-18} \ J = -2.422 \times 10^{-19} \ J$.

(A) The energy of an electronic transition is given by $E = \frac{hc}{\lambda}$,which implies $E \propto \frac{1}{\lambda}$.
According to the Bohr model,the energy of an orbit is $E_n \propto Z^2$. For the same transition (same $n_1$ and $n_2$),the energy difference $\Delta E \propto Z^2$.
Therefore,$\frac{1}{\lambda} \propto Z^2$,or $\lambda \propto \frac{1}{Z^2}$.
For hydrogen $(H)$,$Z = 1$ and $\lambda_H = 912 \mathring{A}$.
For lithium ion $(Li^{2+})$,$Z = 3$.
Using the relation $\frac{\lambda_{Li}}{\lambda_H} = \frac{Z_H^2}{Z_{Li}^2}$,we get:
$\lambda_{Li} = \lambda_H \times \frac{1^2}{3^2} = \frac{912}{9} \mathring{A} = 101.3 \mathring{A}$.

(B) The energy of an electron in a hydrogen-like species is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
For the same excited state,$n$ is constant,so $E \propto Z^2$.
Given the ratio of energies $\frac{E_H}{E_{Li^{2+}}} = \frac{1}{9}$,we have $\frac{Z_H^2}{Z_{Li^{2+}}^2} = \frac{1^2}{3^2} = \frac{1}{9}$,which is consistent with the given data.
The radius of an electron in a hydrogen-like species is given by $r_n = 0.529 \times \frac{n^2}{Z} \mathring{A} $.
For the same excited state,$n$ is constant,so $r \propto \frac{1}{Z}$.
Therefore,the ratio of radii is $\frac{r_H}{r_{Li^{2+}}} = \frac{Z_{Li^{2+}}}{Z_H} = \frac{3}{1} = 3: 1$.

(C) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$.
For the first orbit,$n = 1$,so $r_1 \propto \frac{1}{Z}$.
For $He^{+}$ $(Z = 2)$,$r_{He^+} \propto \frac{1}{2}$.
For $Li^{2+}$ $(Z = 3)$,$r_{Li^{2+}} \propto \frac{1}{3}$.
For $Be^{3+}$ $(Z = 4)$,$r_{Be^{3+}} \propto \frac{1}{4}$.
The ratio is $\frac{1}{2} : \frac{1}{3} : \frac{1}{4}$.
To simplify,multiply by the least common multiple of $2, 3, 4$,which is $12$.
Ratio $= (\frac{1}{2} \times 12) : (\frac{1}{3} \times 12) : (\frac{1}{4} \times 12) = 6 : 4 : 3$.

(A) The wave number $(\bar{\nu})$ is given by the Rydberg formula: $\bar{\nu} = R \cdot Z^2 \left( \frac{1}{n_{L}^2} - \frac{1}{n_{H}^2} \right)$.
For hydrogen,$Z = 1$.
For the lowest energy transition in the Balmer series,$n_{L} = 2$ and $n_{H} = 3$:
$\bar{\nu}_{Balmer} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9-4}{36} \right) = \frac{5}{36} R$.
For the lowest energy transition in the Lyman series,$n_{L} = 1$ and $n_{H} = 2$:
$\bar{\nu}_{Lyman} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R$.
The ratio of the wave number of the Balmer series to the Lyman series is:
$\frac{\bar{\nu}_{Balmer}}{\bar{\nu}_{Lyman}} = \frac{\frac{5}{36} R}{\frac{3}{4} R} = \frac{5}{36} \times \frac{4}{3} = \frac{5}{9 \times 3} = \frac{5}{27}$.
Thus,the ratio is $5 : 27$.

(B) According to Einstein's photoelectric equation,$K.E. = h(v - v_0)$.
For frequency $v_1 = 1.6 \times 10^{16} \ Hz$,$K.E._1 = h(1.6 \times 10^{16} - v_0) \longrightarrow \text{Eq. } (i)$.
For frequency $v_2 = 1.0 \times 10^{16} \ Hz$,$K.E._2 = h(1.0 \times 10^{16} - v_0) \longrightarrow \text{Eq. } (ii)$.
Given that $K.E._1 = 2 \times K.E._2$,we have:
$h(1.6 \times 10^{16} - v_0) = 2 \times h(1.0 \times 10^{16} - v_0)$.
$1.6 \times 10^{16} - v_0 = 2.0 \times 10^{16} - 2v_0$.
$2v_0 - v_0 = 2.0 \times 10^{16} - 1.6 \times 10^{16}$.
$v_0 = 0.4 \times 10^{16} \ Hz = 4 \times 10^{15} \ Hz$.

(B) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.
Given that $\Delta x = \Delta p$,we substitute this into the equation: $(\Delta p)^2 = \frac{h}{4\pi}$.
Since $\Delta p = m \Delta v$,we have $(m \Delta v)^2 = \frac{h}{4\pi}$.
Expanding this gives $m^2 \Delta v^2 = \frac{h}{4\pi}$.
Solving for $\Delta v^2$,we get $\Delta v^2 = \frac{h}{4\pi m^2}$.
Taking the square root of both sides,we obtain $\Delta v = \frac{1}{m} \sqrt{\frac{h}{4\pi}}$.

(B) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$.
Since $p = \sqrt{2mK}$,where $m$ is mass and $K$ is kinetic energy,the ratio of wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{p_2}{p_1} = \sqrt{\frac{2m_2 K_2}{2m_1 K_1}}$.
Given $\frac{m_1}{m_2} = \frac{1}{3}$ and $\frac{K_1}{K_2} = \frac{2}{1}$,we have $\frac{m_2}{m_1} = 3$ and $\frac{K_2}{K_1} = \frac{1}{2}$.
Substituting these values: $\frac{\lambda_1}{\lambda_2} = \sqrt{3 \times \frac{1}{2}} = \sqrt{\frac{3}{2}} = \sqrt{3}:\sqrt{2}$.

(C) Given: Work function $(W_0) = 4.80 \ eV$ and Kinetic energy $(KE) = 0.20 \ eV$.
According to the photoelectric effect equation:
Total energy $(E_T) = W_0 + KE = 4.80 \ eV + 0.20 \ eV = 5.0 \ eV$.
Convert energy to Joules: $E_T = 5.0 \times 1.602 \times 10^{-19} \ J = 8.01 \times 10^{-19} \ J$.
Using the relation $E_T = h\nu$,where $h = 6.626 \times 10^{-34} \ J \cdot s$:
$\nu = \frac{E_T}{h} = \frac{8.01 \times 10^{-19} \ J}{6.626 \times 10^{-34} \ J \cdot s} \approx 1.21 \times 10^{15} \ Hz$.

(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Given $\lambda = 450 \ nm = 450 \times 10^{-9} \ m$.
$E = \frac{6.626 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m/s}{450 \times 10^{-9} \ m} \approx 4.417 \times 10^{-19} \ J$.
Converting to $eV$: $E = \frac{4.417 \times 10^{-19} \ J}{1.602 \times 10^{-19} \ J/eV} \approx 2.76 \ eV$.
Photoelectric effect occurs if the energy of the incident photon is greater than or equal to the work function $(E \ge W_0)$.
Metals with $W_0 < 2.76 \ eV$ are $K (2.25 \ eV), Na (2.30 \ eV)$,and $Li (2.42 \ eV)$. These will undergo the photoelectric effect.
Metals with $W_0 > 2.76 \ eV$ are $Mg (3.70 \ eV)$ and $Cu (4.80 \ eV)$. These will not undergo the photoelectric effect.
The number of such metals is $2$.

(C) The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2}mv^2$,so $v = \sqrt{\frac{2 \times K.E.}{m}}$.
Substituting the values: $v = \sqrt{\frac{2 \times 8.0 \times 10^{-25}}{9.0 \times 10^{-31}}} = \sqrt{\frac{16}{9} \times 10^6} = \frac{4}{3} \times 10^3 \ m/s$.
The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
Substituting $h = 6.626 \times 10^{-34} \ J \cdot s$,$m = 9.0 \times 10^{-31} \ kg$,and $v = \frac{4}{3} \times 10^3 \ m/s$:
$\lambda = \frac{6.626 \times 10^{-34}}{9.0 \times 10^{-31} \times (4/3) \times 10^3} = \frac{6.626 \times 10^{-34}}{12 \times 10^{-28}} = 0.55216 \times 10^{-6} \ m$.
Converting to $nm$: $\lambda = 0.55216 \times 10^{-6} \times 10^9 \ nm = 552.16 \ nm \approx 552.2 \ nm$.

(B) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \ge \frac{h}{4 \pi}$.
Given that the uncertainty in position $(\Delta x)$ is equal to the uncertainty in momentum $(\Delta p)$: $\Delta x = \Delta p$.
Substituting this into the equation: $(\Delta p)^2 = \frac{h}{4 \pi}$.
Therefore,$\Delta p = \sqrt{\frac{h}{4 \pi}}$.
Since momentum uncertainty is related to velocity uncertainty by $\Delta p = m \Delta v$,we have $m \Delta v = \sqrt{\frac{h}{4 \pi}}$.
Solving for $\Delta v$: $\Delta v = \frac{1}{m} \sqrt{\frac{h}{4 \pi}} = \frac{1}{m} \cdot \frac{1}{2} \sqrt{\frac{h}{\pi}} = \frac{1}{2m} \sqrt{\frac{h}{\pi}}$.

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2 m K E}}$.
Given the mass ratio $\frac{m_1}{m_2} = \frac{1}{3}$ and kinetic energy ratio $\frac{K E_1}{K E_2} = \frac{2}{1}$.
The ratio of wavelengths is $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2 \times K E_2}{m_1 \times K E_1}}$.
Substituting the values: $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{3 \times 1}{1 \times 2}} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}}$.
Thus,the ratio is $\sqrt{3} : \sqrt{2}$.

(C) The electronic configuration of xenon $(Xe)$ with atomic number $54$ is: $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^{10}, 4s^2, 4p^6, 4d^{10}, 5s^2, 5p^6$.
Counting the orbitals:
$1s$ ($1$ orbital),$2s$ ($1$ orbital),$2p$ ($3$ orbitals),$3s$ ($1$ orbital),$3p$ ($3$ orbitals),$3d$ ($5$ orbitals),$4s$ ($1$ orbital),$4p$ ($3$ orbitals),$4d$ ($5$ orbitals),$5s$ ($1$ orbital),$5p$ ($3$ orbitals).
Total number of orbitals = $1 + 1 + 3 + 1 + 3 + 5 + 1 + 3 + 5 + 1 + 3 = 27$.

(C) Physical adsorption is an exothermic process. According to Le Chatelier's principle,for an exothermic process,the extent of adsorption decreases with an increase in temperature.
From the given graph,at a constant pressure $p$,the extent of adsorption $\frac{x}{m}$ follows the order: $T_3 > T_2 > T_1$.
Since adsorption is inversely proportional to temperature for physical adsorption,the relationship between the temperatures must be $T_1 > T_2 > T_3$,which is equivalent to $T_3 < T_2 < T_1$.

(B) According to Newton's law of cooling, the rate of cooling is proportional to the heat lost per unit time. Assuming the rate of heat loss $(dQ/dt)$ is constant for the same temperature range:
Heat lost $Q = (w + m) C_V \Delta T$
where $w$ is the water equivalent of the calorimeter, $m$ is the mass of water, $C_V$ is the specific heat of water, and $\Delta T = 62^{\circ} C - 58^{\circ} C = 4^{\circ} C$.
Since the rate of cooling is constant, the time taken $t \propto Q$.
Case $I$: $t_1 = 9 \ \text{minutes}$, $m_1 = 75 \ g$
$9 \propto (w + 75) C_V \times 4^{\circ} \quad \dots (i)$
Case $II$: $t_2 = 12 \ \text{minutes}$, $m_2 = 105 \ g$
$12 \propto (w + 105) C_V \times 4^{\circ} \quad \dots (ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{9}{12} = \frac{(w + 75) C_V \times 4^{\circ}}{(w + 105) C_V \times 4^{\circ}}$
$\frac{3}{4} = \frac{w + 75}{w + 105}$
$3(w + 105) = 4(w + 75)$
$3w + 315 = 4w + 300$
$w = 315 - 300 = 15 \ g$
Thus, the water equivalent of the calorimeter is $15 \ g$.

(A) The amount of heat $Q$ transmitted through a rod in time $t$ is given by the formula:
$Q = \frac{k A t (T_1 - T_2)}{d}$
where $k$ is the thermal conductivity,$A$ is the cross-sectional area,$d$ is the length of the rod,and $(T_1 - T_2)$ is the temperature difference.
Since the rods have the same dimensions,$A$ and $d$ are constant for all rods.
At the junction $P$,let the temperature be $T$. According to the principle of conservation of energy in the steady state,the heat flowing into the junction must equal the heat flowing out of the junction.
Heat flowing from the $100^{\circ} C$ end to $P$ equals the sum of heat flowing from $P$ to the $50^{\circ} C$ end and from $P$ to the $0^{\circ} C$ end:
$\frac{3K A (100 - T) t}{d} = \frac{2K A (T - 50) t}{d} + \frac{K A (T - 0) t}{d}$
Canceling the common terms $\frac{K A t}{d}$ from both sides:
$3(100 - T) = 2(T - 50) + (T - 0)$
$300 - 3T = 2T - 100 + T$
$300 - 3T = 3T - 100$
$400 = 6T$
$T = \frac{400}{6} = \frac{200}{3}{ }^{\circ} C$

(A) The correct matches are as follows:
$(A)$ At constant volume,the heat exchanged is equal to the change in internal energy: $q_V = \Delta U$ $(III)$.
$(B)$ For an isothermal irreversible process,the work done is given by: $W = -p_{ex} (V_f - V_i)$ $(IV)$.
$(C)$ For an isothermal reversible process,the work done is given by: $W = -2.303 nRT \log \frac{V_f}{V_i}$ $(I)$.
$(D)$ For an adiabatic process,$q = 0$,so according to the first law of thermodynamics $(\Delta U = q + W)$,we have $W_{adiabatic} = \Delta U$ $(II)$.
Therefore,the correct sequence is $A-III, B-IV, C-I, D-II$.