AP EAMCET 2020 Chemistry Question Paper with Answer and Solution

(C) The given equation is $x^2-4xy-y^2+6x+2y+k=0$.
Comparing this with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get:
$a=1, h=-2, b=-1, g=3, f=1, c=k$.
The condition for the equation to represent a pair of straight lines is $\Delta = abc+2fgh-af^2-bg^2-ch^2=0$.
Substituting the values:
$(1)(-1)(k) + 2(1)(3)(-2) - (1)(1)^2 - (-1)(3)^2 - k(-2)^2 = 0$.
$-k - 12 - 1 + 9 - 4k = 0$.
$-5k - 4 = 0$.
$5k = -4$.
$k = -\frac{4}{5}$.
Thus,the correct option is $C$.

(C) The equations of the lines passing through the origin with slopes $m_1 = \frac{2}{3}$ and $m_2 = -\frac{2}{3}$ are given by $y = m_1 x$ and $y = m_2 x$.
Substituting the slopes,we get $y = \frac{2}{3}x \Rightarrow 2x - 3y = 0$ and $y = -\frac{2}{3}x \Rightarrow 2x + 3y = 0$.
The combined equation is the product of these two linear equations:
$(2x - 3y)(2x + 3y) = 0$.
Using the identity $(a-b)(a+b) = a^2 - b^2$,we get:
$(2x)^2 - (3y)^2 = 0 \Rightarrow 4x^2 - 9y^2 = 0$.

(D) Let $f(x, y) = x^2+xy+2y^2-3x+2y+4=0$. The point of intersection $(x, y)$ of the pair of lines is obtained by solving the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.
$\frac{\partial f}{\partial x} = 2x + y - 3 = 0 \quad \dots (i)$
$\frac{\partial f}{\partial y} = x + 4y + 2 = 0 \quad \dots (ii)$
Solving these linear equations:
From $(i)$,$y = 3 - 2x$.
Substituting in $(ii)$: $x + 4(3 - 2x) + 2 = 0$
$x + 12 - 8x + 2 = 0$
$-7x + 14 = 0 \implies x = 2$.
Substituting $x = 2$ in $(i)$: $2(2) + y - 3 = 0 \implies 4 + y - 3 = 0 \implies y = -1$.
Thus,the point of intersection is $(2, -1)$.
Therefore,option $D$ is correct.

(C) The equation $x^2+4xy+y^2=0$ represents a pair of lines passing through the origin. Let these lines be $y = m_1x$ and $y = m_2x$.
Comparing with $ax^2+2hxy+by^2=0$,we have $a=1, h=2, b=1$.
The slopes $m_1, m_2$ satisfy $m_1+m_2 = -2h/b = -4$ and $m_1m_2 = a/b = 1$.
The angle $\theta$ between these lines is given by $\tan \theta = |\frac{2\sqrt{h^2-ab}}{a+b}| = |\frac{2\sqrt{4-1}}{1+1}| = \sqrt{3}$.
Thus,$\theta = 60^\circ$.
Since the lines pass through the origin and the angle between them is $60^\circ$,the triangle formed by these two lines and the line $x-y=4$ is an equilateral triangle because the line $x-y=4$ makes an angle of $45^\circ$ with the axes,but checking the intersection points shows the triangle is equilateral.

(C) The given equation is $ax^2 + 4xy + y^2 = 0$. Let the slopes of the two lines be $m$ and $3m$.
Comparing the equation with $Ax^2 + 2Hxy + By^2 = 0$,we have $A = a$,$2H = 4$ (so $H = 2$),and $B = 1$.
The sum of the slopes is $m + 3m = -\frac{2H}{B} = -\frac{4}{1} = -4$.
$4m = -4 \Rightarrow m = -1$.
The product of the slopes is $m(3m) = \frac{A}{B} = \frac{a}{1} = a$.
$3m^2 = a$.
Substituting $m = -1$,we get $3(-1)^2 = a$,which implies $a = 3$.
Thus,the correct option is $C$.

(D) Given,the equations of the circles are:
$S: x^2+y^2+6x-2y+k=0$
$S': x^2+y^2+2x-6y-15=0$
The circle $S$ bisects the circumference of circle $S'$,which means the common chord of the two circles passes through the center of circle $S'$.
The equation of the common chord is given by $S - S' = 0$:
$(x^2+y^2+6x-2y+k) - (x^2+y^2+2x-6y-15) = 0$
$4x + 4y + k + 15 = 0$ $(i)$
The center of circle $S'$ is $(-g, -f)$,where $2g=2$ and $2f=-6$,so the center is $(-1, 3)$.
Since the common chord passes through the center $(-1, 3)$,we substitute these coordinates into equation $(i)$:
$4(-1) + 4(3) + k + 15 = 0$
$-4 + 12 + k + 15 = 0$
$8 + k + 15 = 0$
$k + 23 = 0$
$k = -23$
Thus,option $D$ is correct.

(C) The equation of any tangent to the parabola $y^2=4x$ is $y=mx+\frac{1}{m}$.
Since the tangents pass through the point $(1,4)$,we have $4=m(1)+\frac{1}{m}$.
Multiplying by $m$,we get $m^2-4m+1=0$.
Let the slopes of the two tangents be $m_1$ and $m_2$. Then $m_1+m_2=4$ and $m_1m_2=1$.
The difference between the slopes is $|m_1-m_2| = \sqrt{(m_1+m_2)^2-4m_1m_2} = \sqrt{16-4} = \sqrt{12} = 2\sqrt{3}$.
The angle $\theta$ between the tangents is given by $\tan \theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$.
Substituting the values,$\tan \theta = \left|\frac{2\sqrt{3}}{1+1}\right| = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
Therefore,$\theta = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.

(D) matrix has no inverse if and only if its determinant is equal to $0$.
Let $A = \left[\begin{array}{rrr}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]$.
We set $|A| = 0$:
$|A| = 1(x - (-1)) - (-1)(1 - x) + x(-1 - x^2) = 0$
$|A| = 1(x + 1) + 1(1 - x) + x(-1 - x^2) = 0$
$|A| = x + 1 + 1 - x - x - x^3 = 0$
$|A| = -x^3 - x + 2 = 0$
$x^3 + x - 2 = 0$
By inspection,if $x = 1$,then $1^3 + 1 - 2 = 0$,which satisfies the equation.
Alternatively,if $x = 1$,the matrix becomes $\left[\begin{array}{rrr}1 & -1 & 1 \\ 1 & 1 & 1 \\ 1 & -1 & 1\end{array}\right]$.
Since the first and third columns are identical,the determinant is $0$.
Thus,the real value of $x$ is $1$.

(C) To check for injectivity,let $f(x_1) = f(x_2)$.
$\frac{x_1}{\sqrt{1+x_1^2}} = \frac{x_2}{\sqrt{1+x_2^2}}$
Squaring both sides: $\frac{x_1^2}{1+x_1^2} = \frac{x_2^2}{1+x_2^2}$
$x_1^2(1+x_2^2) = x_2^2(1+x_1^2)$
$x_1^2 + x_1^2x_2^2 = x_2^2 + x_1^2x_2^2$
$x_1^2 = x_2^2 \Rightarrow x_1 = x_2$ (since the function is strictly increasing,$x_1 = x_2$).
Thus,$f$ is injective.
To check for surjectivity,let $y = \frac{x}{\sqrt{1+x^2}}$.
$y^2 = \frac{x^2}{1+x^2} \Rightarrow y^2(1+x^2) = x^2 \Rightarrow y^2 = x^2(1-y^2) \Rightarrow x^2 = \frac{y^2}{1-y^2}$.
For $x$ to be a real number,$1-y^2 > 0$,which means $y^2 < 1$,so $-1 < y < 1$.
The range of $f$ is $(-1, 1)$,which is not equal to the codomain $R$.
Therefore,$f$ is not surjective.

(A) Given that,$x = e^{y+e^{y+e^{y+\ldots}}}$.
Since the exponent repeats infinitely,we can write $x = e^{y+x}$.
Taking the natural logarithm on both sides,we get $\log_e x = y+x$.
Rearranging for $y$,we have $y = \log_e x - x$.
Differentiating both sides with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(\log_e x) - \frac{d}{dx}(x)$.
Thus,$\frac{dy}{dx} = \frac{1}{x} - 1 = \frac{1-x}{x}$.
Therefore,option $(A)$ is correct.

(C) disproportionation reaction is a type of redox reaction in which the same element is simultaneously oxidized and reduced.
For an element to undergo disproportionation,it must be in an intermediate oxidation state.
In $ClO_4^{-}$,the oxidation state of $Cl$ is calculated as: $x + 4(-2) = -1$,which gives $x = +7$.
Since $+7$ is the maximum possible oxidation state for chlorine (as it has $7$ valence electrons),it cannot be further oxidized.
Therefore,$ClO_4^{-}$ cannot undergo a disproportionation reaction.

(D) Disproportionation is a specific type of redox reaction in which a species is simultaneously reduced and oxidized to form two different products.
In a disproportionation reaction,the central atom must be in an intermediate oxidation state so that it can both increase and decrease its oxidation number.
The oxidation state of $Cl$ in the given species is:
$ClO^{-}$: $+1$
$ClO_2^{-}$: $+3$
$ClO_3^{-}$: $+5$
$ClO_4^{-}$: $+7$
Since $Cl$ in $ClO_4^{-}$ is already in its maximum oxidation state of $+7$,it cannot be further oxidized.
Therefore,$ClO_4^{-}$ cannot undergo a disproportionation reaction.
Hence,the correct option is $(D)$.

(C) In the reaction $4 KClO_3 \longrightarrow 3 KClO_4 + KCl$,the oxidation state of chlorine in $KClO_3$ is $+5$.
In $KClO_4$,the oxidation state of chlorine is $+7$ (oxidation).
In $KCl$,the oxidation state of chlorine is $-1$ (reduction).
Since the same element (chlorine) is simultaneously oxidized and reduced,this is a disproportionation reaction.
Hence,the correct option is $(C)$.

(D) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
For an element to undergo disproportionation,it must exist in an intermediate oxidation state,meaning it should be capable of both increasing and decreasing its oxidation number.
In $ClO_4^-$,the oxidation state of chlorine is $+7$.
Since chlorine is in its maximum possible oxidation state $(+7)$,it cannot be further oxidized.
Therefore,$ClO_4^-$ cannot undergo a disproportionation reaction.

(A) The oxidation number of hydrogen is usually $+1$ in compounds with more electronegative atoms (as in $H_2O$).
However,in metal hydrides like lithium hydride $(LiH)$,the oxidation state of the hydrogen atom is $-1$.
Therefore,the statement that the oxidation number of hydrogen is always $+1$ is incorrect.

(C) The redox reaction is: $a C_2O_4^{2-} + b MnO_4^- + H^+ \longrightarrow p CO_2 + q Mn^{2+} + H_2O$
Step $1$: Determine the $n$-factors.
For $C_2O_4^{2-}$,the oxidation state of $C$ changes from $+3$ to $+4$. The $n$-factor is $(4-3) \times 2 = 2$.
For $MnO_4^-$,the oxidation state of $Mn$ changes from $+7$ to $+2$. The $n$-factor is $(7-2) \times 1 = 5$.
Step $2$: Balance the $n$-factors by cross-multiplying.
We take $5$ moles of oxalate $(a=5)$ and $2$ moles of permanganate $(b=2)$.
Step $3$: Balance the atoms.
For $C$-atoms: $2a = p \implies p = 2 \times 5 = 10$.
For $Mn$-atoms: $b = q \implies q = 2$.
For $O$-atoms: $5(4) + 2(4) = 10(2) + 2(1) + H_2O \implies 28 = 22 + H_2O \implies H_2O = 6$.
For $H$-atoms: $H^+ = 2 \times 6 = 12$.
The balanced equation is $5C_2O_4^{2-} + 2MnO_4^- + 16H^+ \longrightarrow 10CO_2 + 2Mn^{2+} + 8H_2O$.
Thus,the coefficient of $CO_2$ is $10$.

(B) The molar mass of $MnO_4^{-}$ is calculated as: $55 + (4 \times 16) = 55 + 64 = 119 \ g/mol$.
In an acidic medium,the reduction reaction is: $MnO_4^{-} + 8H^{+} + 5e^{-} \longrightarrow Mn^{2+} + 4H_2O$.
Here,the change in oxidation state of $Mn$ is from $+7$ to $+2$,so the $n$-factor is $5$.
$\text{Equivalent weight} = \frac{\text{Molar mass}}{n\text{-factor}} = \frac{119}{5} = 23.8$.
Hence,the correct option is $B$.

(D) The balanced chemical equation for the reaction in acidic medium is: $10I^{-} + 2MnO_4^{-} + 16H^{+} \rightarrow 5I_2 + 2Mn^{2+} + 8H_2O$.
In $KMnO_4$,the oxidation state of $Mn$ is calculated as: $x + 1(-2) = -1 \Rightarrow x = +7$.
In the product $Mn^{2+}$,the oxidation state of $Mn$ is $+2$.
The change in the oxidation number of manganese is $|(+2) - (+7)| = 5$ units.

(A) Among alkali metals (group-$1$),$Cs$ is the most reactive because it has the lowest ionisation energy ($IE$,or $\Delta_i H_1$).
Note: $Fr$ is not considered as it is radioactive.

(C) $Li$ has the highest hydration enthalpy among alkali metals,which accounts for its high negative $E^{\circ}$ value and its high reducing power in aqueous solution.
Hence,the correct option is $(C)$.

(A) When alkali metals like sodium dissolve in liquid ammonia,they undergo the following reaction: $Na + (x+y)NH_3 \rightarrow [Na(NH_3)_x]^+ + [e(NH_3)_y]^-$.
The deep blue colour of the solution is primarily due to the presence of ammoniated electrons,which absorb energy in the visible region of the spectrum to promote electrons to higher energy levels.

(A) $Li^{+}$ (group-$1$) and $Mg^{2+}$ (group-$2$) exhibit a diagonal relationship in the periodic table,resulting in similar ionic radii.
Both $Li$ and $Mg$ react with oxygen to form only monoxides ($Li_2O$ and $MgO$ respectively) and react directly with nitrogen to form nitrides ($Li_3N$ and $Mg_3N_2$ respectively).
Therefore,the metals $A$ and $B$ are $Li$ and $Mg$.

(C) $Ca(OH)_2$ (calcium hydroxide) is used in the Solvay process for the preparation of washing soda $(Na_2CO_3 \cdot 10H_2O)$ to recover ammonia $(NH_3)$ from ammonium chloride $(NH_4Cl)$.
$2NH_4Cl + Ca(OH)_2 \longrightarrow 2NH_3 + CaCl_2 + 2H_2O$
When $CO_2$ is passed through an aqueous solution of $Ca(OH)_2$ (lime water),it forms $CaCO_3$,which makes the solution milky.
$Ca(OH)_2 + CO_2 \longrightarrow CaCO_3 \downarrow + H_2O$
$Ca(OH)_2$ is also used in white washing due to its disinfectant nature.
Therefore,the correct option is $C$.

(C) Statement $(i)$ is correct: Beryllium is not readily attacked by acids due to the formation of a protective oxide layer on its surface.
Statement $(ii)$ is correct: $BeO$ is amphoteric in nature,meaning it reacts with both acids and bases.
Statement $(iii)$ is correct: Due to its small size and availability of vacant $2p$ orbitals,Beryllium can exhibit a coordination number greater than $4$ (e.g.,in $[Be(H_2O)_4]^{2+}$ or complex fluorides).
Statement $(iv)$ is incorrect: $BeO$ is amphoteric,not purely acidic.
Therefore,statements $(i)$,$(ii)$,and $(iii)$ are correct.

(B) Since the vectors are coplanar,their scalar triple product is zero:
$\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0$
Applying row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$\begin{vmatrix} a & 1 & 1 \\ 1-a & b-1 & 0 \\ 1-a & 0 & c-1 \end{vmatrix} = 0$
Expanding along the first row:
$a((b-1)(c-1) - 0) - 1((1-a)(c-1) - 0) + 1(0 - (1-a)(b-1)) = 0$
$a(b-1)(c-1) - (1-a)(c-1) - (1-a)(b-1) = 0$
Divide the entire equation by $(1-a)(1-b)(1-c)$ (noting $a, b, c \neq 1$):
$\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(b-1)}{(1-a)(1-b)(1-c)} = 0$
$\frac{-a}{(1-a)} - \frac{1}{(1-b)} - \frac{1}{(1-c)} = 0$
Since $\frac{-a}{1-a} = \frac{1-a-1}{1-a} = 1 - \frac{1}{1-a}$,we substitute:
$1 - \frac{1}{1-a} - \frac{1}{1-b} - \frac{1}{1-c} = 0$
Therefore,$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$.

(C) Given the equation $|u-v| = ||u|-|v||$.
Squaring both sides,we get:
$|u-v|^2 = (||u|-|v||)^2$
$|u|^2 + |v|^2 - 2(u \cdot v) = |u|^2 + |v|^2 - 2|u||v|$
$-2(u \cdot v) = -2|u||v|$
$u \cdot v = |u||v|$
Since $u \cdot v = |u||v| \cos \theta$,we have $|u||v| \cos \theta = |u||v|$.
This implies $\cos \theta = 1$,which means $\theta = 0$.
Therefore,$u$ and $v$ have the same direction.

(A) According to Heisenberg's uncertainty principle,it is impossible to determine the exact position and exact momentum of a microscopic particle like an electron simultaneously. $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$.
This principle is significant only for microscopic particles. For macroscopic objects,the uncertainty is negligible and practically non-existent.

(B) The covalent radius is defined as half the distance between the nuclei of two covalently bonded atoms. Since there is orbital overlap,the distance is the smallest.
In metallic bonding,atoms are in contact but there is no orbital overlap,making the metallic radius larger than the covalent radius.
Van der Waals radius corresponds to the distance between non-bonded atoms in adjacent molecules,which is the largest due to the absence of any chemical bond.
Therefore,the correct order is: $Covalent \ radius < Metallic \ radius < Van \ der \ Waals \ radius$.

(D) The graph represents the cooling curve of a mixture of two liquids $A$ and $B$.
At point $B$ $(-25^{\circ} C)$,liquid $B$ starts to solidify. Along the curve $BC$,liquid $B$ is solidifying,so two phases (liquid $B$ and solid $B$) co-exist along with liquid $A$.
At point $C$ $(-78^{\circ} C)$,liquid $A$ also starts to solidify.
Along the line $CD$,both liquid $A$ and liquid $B$ are undergoing solidification.
Therefore,the phases present are: liquid $A$,solid $A$,liquid $B$,and solid $B$.
Thus,$4$ different phases co-exist along the line $CD$.

(D) Assertion: The statement is incorrect because the standard boiling point (boiling point at $1 \ bar$ pressure) of a liquid is slightly lower than the normal boiling point (boiling point at $1 \ atm$ pressure).
Reason: The statement is correct because $1 \ atm = 1.01325 \ bar$,which means $1 \ bar$ pressure is slightly lower than $1 \ atm$ pressure.
Since the Assertion is incorrect and the Reason is correct,the correct option is $D$.

(B) The temperature at which the vapor pressure of a liquid becomes equal to the external pressure is defined as the boiling point of the liquid.
At a standard atmospheric pressure of $1.013 \ bar$ $(1 \ atm)$,this temperature is specifically known as the normal boiling point.

(B) One mole of gas at $STP$ occupies $22.4 \ L$.
Thus,$5.6 \ L$ of gas at $STP$ contains $n = \frac{5.6}{22.4} = 0.25 \ \text{moles}$.
Number of moles = $\frac{\text{weight in gram}}{\text{molecular weight (M)}}$.
$0.25 = \frac{7.5}{M}$.
$M = \frac{7.5}{0.25} = 30 \ \text{g/mol}$.
The molar mass of $NO$ is $14 + 16 = 30 \ \text{g/mol}$.
Thus,the gas is $NO$.
Hence,the correct option is $B$.

(C) The balanced chemical equation for the combustion of methane is:
$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
Molar mass of $CH_4 = 12 + 4(1) = 16 \text{ g/mol}$
Molar mass of $CO_2 = 12 + 2(16) = 44 \text{ g/mol}$
From the stoichiometry of the reaction,$16 \text{ g}$ of $CH_4$ produces $44 \text{ g}$ of $CO_2$.
Therefore,$32 \text{ g}$ of $CH_4$ will produce:
$\text{Amount of } CO_2 = \frac{44 \times 32}{16} = 88 \text{ g}$

(B) The molecular weight of the metal chloride is $2 \times \text{vapour density} = 2 \times 83 = 166$.
Let the valency of the metal be $n$. The formula of the metal chloride is $MCl_n$.
Molecular weight $= \text{Atomic weight of metal} + n \times 35.5$.
Since $\text{Atomic weight} = n \times \text{Equivalent weight} = n \times 6$,we have:
$166 = 6n + 35.5n$.
$166 = 41.5n$.
$n = \frac{166}{41.5} = 4$.
Atomic weight of metal $= n \times \text{Equivalent weight} = 4 \times 6 = 24$.
Hence,the correct option is $B$.

(C) Step $1$: Calculate the moles of reactants.
Molar mass of $Cu \approx 63.5 \ g/mol$,Molar mass of $I_2 \approx 253.8 \ g/mol$.
Moles of $Cu = \frac{10}{63.5} \approx 0.157 \ mol$.
Moles of $I_2 = \frac{10}{253.8} \approx 0.0394 \ mol$.
Step $2$: Identify the limiting reagent.
According to the equation $2Cu + I_2 \longrightarrow 2CuI$,$1 \ mol$ of $I_2$ requires $2 \ mol$ of $Cu$.
For $0.0394 \ mol$ of $I_2$,we need $2 \times 0.0394 = 0.0788 \ mol$ of $Cu$.
Since we have $0.157 \ mol$ of $Cu$ (which is $> 0.0788 \ mol$),$I_2$ is the limiting reagent.
Step $3$: Calculate the yield of $CuI$.
Moles of $CuI$ produced $= 2 \times \text{moles of } I_2 = 2 \times 0.0394 = 0.0788 \ mol$.
Molar mass of $CuI = 63.5 + 126.9 = 190.4 \ g/mol$.
Mass of $CuI = 0.0788 \ mol \times 190.4 \ g/mol \approx 15 \ g$.
Thus,the correct option is $C$.

(A) The balanced chemical equation is: $3 NO_2 + H_2O \longrightarrow 2 HNO_3 + NO$.
From the stoichiometry of the reaction,$3$ moles of $NO_2$ produce $2$ moles of $HNO_3$.
To produce $4$ moles of $HNO_3$,we need:
$\text{Moles of } NO_2 = \frac{3}{2} \times 4 = 6 \text{ moles}$.
The molar mass of $NO_2 = 14 + (2 \times 16) = 46 \ g/mol$.
Therefore,the mass of $NO_2$ required $= 6 \text{ moles} \times 46 \ g/mol = 276 \ g$.
Hence,the correct option is $A$.

(C) Given,density $(d)$ of $HCl$ solution $= 1.2 \ g \ mL^{-1}$.
Percentage strength by mass $(w/w)$ $= 40 \%$.
Molar mass of $HCl$ $(M_{HCl})$ $= 1 + 35.5 = 36.5 \ g \ mol^{-1}$.
The formula for molarity is: $\text{Molarity} = \frac{\text{Percentage}(w/w) \times d \times 10}{M_{HCl}}$.
Substituting the values: $\text{Molarity} = \frac{40 \times 1.2 \times 10}{36.5} = \frac{480}{36.5} \approx 13.15 \ M$.
Therefore,the molarity is nearly $13 \ M$.

(D) The estimation of phosphorus in an organic compound is performed by using a magnesia mixture.
$1 \ mol$ of $Mg_2P_2O_7$ contains $2 \ mol$ of phosphorus $(P)$.
Molar mass of $Mg_2P_2O_7 = (2 \times 24.3) + (2 \times 31) + (7 \times 16) = 48.6 + 62 + 112 = 222.6 \ g/mol$.
Mass of phosphorus in $0.22 \ g$ of $Mg_2P_2O_7 = \frac{2 \times 31}{222.6} \times 0.22 \ g = 0.06127 \ g$.
Percentage of phosphorus $= \frac{\text{Mass of } P}{\text{Mass of compound}} \times 100 = \frac{0.06127}{0.12} \times 100 \approx 51.06 \%$.
Using the standard formula: $P \% = \frac{62}{222.6} \times \frac{w_1}{w} \times 100 = \frac{62}{222.6} \times \frac{0.22}{0.12} \times 100 \approx 51.06 \%$.
Rounding to the nearest provided option,the correct answer is $51.20 \%$.

(D) The balanced chemical equation for the reaction is: $2H_2(g) + O_2(g) \rightarrow 2H_2O(g)$.
According to the stoichiometry,$2 \ volumes$ of $H_2$ react with $1 \ volume$ of $O_2$.
Given initial volumes: $V(H_2) = 30 \ mL$ and $V(O_2) = 20 \ mL$.
Since $H_2$ is the limiting reagent,all $30 \ mL$ of $H_2$ will be consumed.
The volume of $O_2$ required to react with $30 \ mL$ of $H_2$ is $30 \ mL \times (1/2) = 15 \ mL$.
Therefore,the leftover volume of $O_2$ is $20 \ mL - 15 \ mL = 5 \ mL$ of $O_2$.

(C) The $n$-factor for $Na_2CO_3$ is $2$ because it acts as a base and can accept $2 \ H^+$ ions.
Normality is related to molarity by the formula: $\text{Normality} = n\text{-factor} \times \text{Molarity}$.
Given: $\text{Normality} = 0.2 \ N$ and $n\text{-factor} = 2$.
Therefore,$\text{Molarity} = \frac{\text{Normality}}{n\text{-factor}} = \frac{0.2}{2} = 0.1 \ M$.
Hence,the correct option is $C$.

(A) According to Graham's Law,the rate of diffusion $(r)$ of a gas is inversely proportional to the square root of its molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.
First,calculate the molar masses of the given molecules:
$CO_2: 12 + 2(16) = 44 \ g/mol$
$SO_2: 32 + 2(16) = 64 \ g/mol$
$SO_3: 32 + 3(16) = 80 \ g/mol$
$PCl_3: 31 + 3(35.5) = 31 + 106.5 = 137.5 \ g/mol$
Since the rate of diffusion is inversely proportional to the square root of the molar mass,the molecule with the lowest molar mass will diffuse the fastest.
The order of molar masses is: $CO_2 (44) < SO_2 (64) < SO_3 (80) < PCl_3 (137.5)$.
Therefore,the order of the rate of diffusion is: $CO_2 > SO_2 > SO_3 > PCl_3$.

(A) Isochores are drawn at constant volume,$\Delta V = 0$.
We get a plot of pressure $(p)$ $vs$ temperature $(T$,in $K)$ as shown in the graph.
For $1 \ mol$ of an ideal gas,$pV = nRT$.
At isochoric condition,$V$ is constant,so $p \propto T$.

(A) $1$ mole of $O_{2(g)}$ at $STP$ contains Avogadro number $(N_A = 6.022 \times 10^{23} \ mol^{-1})$ of $O_2$ molecules.
Since $1$ mole of $O_2$ has a molar mass of $32 \ g/mol$,it corresponds to $32 \ g$ of oxygen gas.
Therefore,the correct statement is that it contains $6.022 \times 10^{23}$ molecules of oxygen.

(B) The density $(d)$ of a gas is given by the formula: $d = \frac{pM}{RT}$
Given:
Pressure $(p)$ = $5 \text{ atm}$
Molar mass of $O_2$ $(M)$ = $32 \text{ g mol}^{-1}$
Gas constant $(R)$ = $0.082 \text{ L atm K}^{-1} \text{ mol}^{-1}$
Temperature $(T)$ = $127 + 273 = 400 \text{ K}$
Substituting the values:
$d = \frac{5 \times 32}{0.082 \times 400}$
$d = \frac{160}{32.8} \approx 4.878 \text{ g L}^{-1}$
Thus,the density is approximately $4.88 \text{ g L}^{-1}$.

(A) Using the ideal gas equation,$PV = nRT$,the temperature $T$ is given by $T = \frac{PV}{nR}$.
Given values are $P = 3.32 \ bar$,$V = 5 \ dm^3 = 5 \ L$,$n = 4 \ mol$,and the gas constant $R = 0.08314 \ bar \ L \ K^{-1} \ mol^{-1}$.
Substituting these values into the equation:
$T = \frac{3.32 \times 5}{4 \times 0.08314} = \frac{16.6}{0.33256} \approx 49.91 \ K$.
Rounding to the nearest whole number,we get $T \approx 50 \ K$.

(A) According to Boyle's law,at constant temperature,$PV = k$.
$1.$ For $y = P$ and $x = V$,the relation $P = k/V$ represents a hyperbola. Thus,$(A)$ matches $2$.
$2.$ For $y = P$ and $x = 1/V$,the relation $P = k(1/V)$ represents a straight line passing through the origin. Thus,$(B)$ matches $3$.
$3.$ For $y = PV$ and $x = P$,the relation $PV = k$ represents a horizontal line parallel to the $x$-axis. Thus,$(C)$ matches $1$.

(B) Given: $n = 4 \ mol$,$V = 5 \ L$,$P = 3.32 \ bar$,$R = 0.083 \ bar \ L \ K^{-1} \ mol^{-1}$.
Using the ideal gas equation: $PV = nRT$.
Rearranging for temperature: $T = \frac{PV}{nR}$.
Substituting the values: $T = \frac{3.32 \times 5}{4 \times 0.083}$.
$T = \frac{16.6}{0.332} = 50 \ K$.
Therefore,the correct option is $B$.

(C) The variation in the volume of a gas with temperature was first studied by Jacques Charles $(1787)$ and is known as Charles's law.
Charles's law states that,"The volume of a given mass of a gas is directly proportional to its absolute temperature at constant pressure."
Mathematically,$V \propto T$ or $\frac{V}{T} = k$ at constant pressure.
Hence,the correct option is $(C)$.

(D) According to Boyle's law,at constant temperature,$P_1V_1 = P_2V_2$.
Let the initial volume be $V_1 = V$.
The new volume $V_2 = V + 0.10V = 1.10V$.
Substituting into the equation: $P_1V = P_2(1.10V)$.
$P_2 = P_1 / 1.10 \approx 0.9091 P_1$.
The change in pressure is $P_2 - P_1 = 0.9091 P_1 - P_1 = -0.0909 P_1$.
This represents a decrease of approximately $9.09 \%$.
Given the options provided,the closest value is a decrease of $10 \%$,which corresponds to option $(D)$.

(B) Charles' law states that the volume of a fixed mass of an ideal gas is directly proportional to its absolute temperature at constant pressure.
At constant pressure,$V \propto T$.
Or,$\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Hence,the correct option is $(B)$.