GSEB 2020 Chemistry Question Paper with Answer and Solution

(A) In a Linear Programming $(LP)$ problem,the feasible region is defined as the set of all points that satisfy all the given constraints simultaneously,including the non-negativity restrictions.
Any point within or on the boundary of this feasible region is called a feasible solution.
Therefore,by definition,a feasible solution must satisfy all the constraints of the problem simultaneously.

(D) catalyst is a substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.
It functions by providing an alternative reaction pathway with a lower $Activation \ energy$ $(E_a)$.
It does not alter the $Gibbs \ energy$ $(\Delta G)$,$Enthalpy$ $(\Delta H)$,or the $Equilibrium \ constant$ $(K_{eq})$ of the reaction.
Therefore,the correct option is $D$.

(C) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.
Taking the natural logarithm on both sides,we get:
$\ln k = \ln A - \frac{E_a}{RT}$
This equation is in the form of a straight line $y = mx + c$,where:
$y = \ln k$
$x = \frac{1}{T}$
$m = -\frac{E_a}{R}$ (slope)
$c = \ln A$ (intercept)
Since the slope is negative $(-\frac{E_a}{R})$,the graph of $\ln k$ versus $\frac{1}{T}$ is a straight line with a negative slope and a positive intercept on the $y$-axis.
This corresponds to the graph shown in option $C$.

(C) The decomposition reaction is: $2NH_{3(g)} \rightarrow N_{2(g)} + 3H_{2(g)}$
For a zero-order reaction,the rate of reaction is equal to the rate constant $k$.
$\text{Rate} = k = 2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
The rate of reaction is expressed as: $\text{Rate} = -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt}$.
Therefore,the rate of production of $N_2$ is equal to the rate of the reaction.
$\frac{d[N_2]}{dt} = \text{Rate} = 2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

(B) The $IUPAC$ name is tetraammineaquachloridocobalt$(III)$ chloride.
$1$. The central metal ion is $Co$ with an oxidation state of $+3$.
$2$. The ligands are four ammine $(NH_3)$,one aqua $(H_2O)$,and one chlorido $(Cl^-)$ group inside the coordination sphere.
$3$. The total charge of the coordination sphere is calculated as: $x + 4(0) + 1(0) + 1(-1) = +3$,which gives $x = +4$. However,the oxidation state is $+3$,so the sphere charge is $3 - 1 = +2$.
$4$. To balance the $+2$ charge of the complex $[Co(NH_3)_4(H_2O)Cl]^{2+}$,two chloride ions $(Cl^-)$ are required outside the coordination sphere.
$5$. Thus,the formula is $[Co(NH_3)_4(H_2O)Cl]Cl_2$.

(C) Amphoteric oxides are those that react with both acids and bases.
$Cr_2O_3$ is a well-known amphoteric oxide.
$V_2O_5$ also exhibits amphoteric character,although it is predominantly acidic,it reacts with strong bases to form vanadates and with strong acids to form oxovanadium ions.
$Mn_2O_7$ and $CrO_3$ are acidic oxides.
$CrO$ is a basic oxide.
$V_2O_4$ is a basic oxide.
Therefore,the correct pair is $V_2O_5$ and $Cr_2O_3$.

(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity of a weak electrolyte at infinite dilution can be calculated using the molar conductivities of strong electrolytes.
For acetic acid $(HAc)$,the expression is:
$\Lambda_{m(HAc)}^0 = \lambda_{H^+} + \lambda_{Ac^-}$
Using strong electrolytes $HCl$,$NaAc$,and $NaCl$:
$\Lambda_{m(HCl)}^0 = \lambda_{H^+} + \lambda_{Cl^-}$
$\Lambda_{m(NaAc)}^0 = \lambda_{Na^+} + \lambda_{Ac^-}$
$\Lambda_{m(NaCl)}^0 = \lambda_{Na^+} + \lambda_{Cl^-}$
Therefore,$\Lambda_{m(HCl)}^0 + \Lambda_{m(NaAc)}^0 - \Lambda_{m(NaCl)}^0 = (\lambda_{H^+} + \lambda_{Cl^-}) + (\lambda_{Na^+} + \lambda_{Ac^-}) - (\lambda_{Na^+} + \lambda_{Cl^-}) = \lambda_{H^+} + \lambda_{Ac^-} = \Lambda_{m(HAc)}^0$.
Thus,the correct option is $B$.

(D) Reducing power is inversely proportional to the standard reduction potential $(E^\circ)$. The lower (more negative) the $E^\circ$ value,the stronger the reducing agent.
Given standard reduction potentials:
$E^\circ_{Ag^{+}/Ag} = 0.80 \ V$
$E^\circ_{Hg^{2+}/Hg} = 0.79 \ V$
$E^\circ_{Cr^{3+}/Cr} = -0.74 \ V$
$E^\circ_{Mg^{2+}/Mg} = -2.37 \ V$
Comparing these values,the increasing order of reducing power is: $Ag < Hg < Cr < Mg$.

(B) When an external opposing potential $(E_{ext})$ is applied to an electrochemical cell and increased slowly,the reaction continues to take place in the same direction until $E_{ext} = E_{cell}$.
If $E_{ext}$ is increased further such that $E_{ext} > E_{cell}$,the reaction reverses and the cell behaves as an electrolytic cell.

(C) Given: $30 \% \ w/w$ $NaOH$ solution.
This means $30 \ g$ of $NaOH$ is present in $100 \ g$ of the solution.
Mass of solute $(NaOH)$ $= 30 \ g$.
Mass of solvent (water) $= 100 \ g - 30 \ g = 70 \ g = 0.07 \ kg$.
Molar mass of $NaOH = 23 + 16 + 1 = 40 \ g/mol$.
Moles of $NaOH = \frac{30 \ g}{40 \ g/mol} = 0.75 \ mol$.
Molality $(m)$ $= \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.75 \ mol}{0.07 \ kg} = 10.71 \ m$.