MHT CET 2021 Chemistry Question Paper with Answer and Solution

(C) According to the Bronsted-Lowry theory,a base is a proton $(H^+)$ acceptor.
In the given options,$NH_{3(aq)}$ acts as a Bronsted-Lowry base because it accepts a proton from the water molecule.
The reaction is: $H_2O_{(\ell)} + NH_{3(aq)} \rightleftharpoons NH_{4(aq)}^+ + OH^-_{(aq)}$

(D) For a weak electrolyte $HA \rightleftharpoons H^{+} + A^{-}$,the dissociation constant $K_{a}$ is given by $K_{a} = \frac{C\alpha^2}{1-\alpha}$.
Assuming $\alpha \ll 1$,we have $K_{a} \approx C\alpha^2$.
From this,$\alpha = \sqrt{\frac{K_{a}}{C}}$.
Since concentration $C = \frac{1}{V}$,where $V$ is the volume,we can substitute $C$ to get $K_{a} = \frac{\alpha^2}{V}$ or $\alpha = \sqrt{K_{a}V}$.
Comparing these with the given options:
Option $A$ is $\alpha = \sqrt{\frac{K_{a}}{C}}$ (Correct).
Option $B$ is $K = \frac{\alpha^2}{V}$ (Correct).
Option $C$ is $K = \alpha^2 C$ (Correct,as $K = C\alpha^2$).
Option $D$ is $\alpha = \sqrt{\frac{K_{a}}{V}}$,which is incorrect because $\alpha = \sqrt{K_{a}V}$.

(B) conjugate acid-base pair differs by a single proton $(H^{+})$.
In the reaction $HCl_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O_{(aq)}^{+} + Cl_{(aq)}^{-}$:
$(1)$ $HCl$ acts as an acid and loses a proton to form its conjugate base,$Cl^{-}$. Thus,$(HCl, Cl^{-})$ is a conjugate acid-base pair.
$(2)$ $H_2O$ acts as a base and accepts a proton to form its conjugate acid,$H_3O^{+}$. Thus,$(H_2O, H_3O^{+})$ is a conjugate acid-base pair.
Comparing this with the options,$(H_2O, H_3O^{+})$ is listed as option $(B)$.

(D) In an aqueous solution at $298 \ K$:
$[H^{+}] \times [OH^{-}] = K_w = 10^{-14} \ (mol \ dm^{-3})^2$
Given $[OH^{-}] = 1 \times 10^{-12} \ mol \ dm^{-3}$
$[H^{+}] = \frac{K_w}{[OH^{-}]}$
$[H^{+}] = \frac{10^{-14}}{1 \times 10^{-12}} = 10^{-2} \ mol \ dm^{-3}$
Therefore,$[H^{+}] = 0.01 \ mol \ dm^{-3}$.

(D) $NaOH$ is a strong base,so it dissociates completely in water: $[NaOH] = [OH^-] = 0.02 \ M = 2 \times 10^{-2} \ M$.
$pOH = -\log[OH^-] = -\log(2 \times 10^{-2}) = 2 - \log 2 = 2 - 0.301 = 1.699 \approx 1.7$.
Using the relation $pH + pOH = 14$ at $25^{\circ}C$:
$pH = 14 - pOH = 14 - 1.7 = 12.3$.

(D) $pH + pOH = 14$ at $298 \ K$.
$pOH = 14 - pH = 14 - 9.95 = 4.05$.
Since $pOH = -\log_{10} \left[OH^{-}\right]$,we have $\left[OH^{-}\right] = 10^{-pOH}$.
$\left[OH^{-}\right] = 10^{-4.05} = 10^{-5 + 0.95} = 10^{-5} \times 10^{0.95}$.
Using $10^{0.95} \approx 8.91$,we get $\left[OH^{-}\right] = 8.91 \times 10^{-5} \ M$.

(A) The formula for $pH$ is given by: $pH = -\log_{10}[H^{+}]$.
Therefore,the concentration of hydrogen ions is: $[H^{+}] = 10^{-pH}$.
Substituting the given value: $[H^{+}] = 10^{-3.6}$.
To solve this,we can write: $[H^{+}] = 10^{-4 + 0.4} = 10^{0.4} \times 10^{-4}$.
Since $10^{0.4} \approx 2.51$,the concentration is $[H^{+}] = 2.51 \times 10^{-4} \ M$.

(D) $H_2SO_4 \longrightarrow 2H^{+} + SO_4^{2-}$
$[H_2SO_4] = 0.005 \ M$
$[H^{+}] = 2 \times 0.005 = 0.01 \ M$
$pH = -\log[H^{+}] = -\log(0.01)$
$pH = 2$

(A) The $pH$ of a salt solution depends on the nature of the acid and base from which it is formed.
$1$. Sodium acetate $(CH_3COONa)$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$. It undergoes anionic hydrolysis to form $OH^-$ ions,making the solution basic $(pH > 7)$.
$2$. Sodium sulphate $(Na_2SO_4)$ is a salt of a strong acid $(H_2SO_4)$ and a strong base $(NaOH)$. It does not undergo hydrolysis,so the solution is neutral $(pH = 7)$.
$3$. Copper sulphate $(CuSO_4)$ is a salt of a strong acid $(H_2SO_4)$ and a weak base $(Cu(OH)_2)$. It undergoes cationic hydrolysis to form $H^+$ ions,making the solution acidic $(pH < 7)$.
$4$. Ammonium chloride $(NH_4Cl)$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$. It undergoes cationic hydrolysis to form $H^+$ ions,making the solution acidic $(pH < 7)$.

(D) The acidity of a salt solution depends on the nature of the acid and base from which it is formed.
$1$. $CH_3COONH_4$ (Ammonium acetate) is a salt of a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$,making it nearly neutral.
$2$. $NH_4CN$ (Ammonium cyanide) is a salt of a weak acid $(HCN)$ and a weak base $(NH_4OH)$,which is basic because $K_b > K_a$.
$3$. $NaCl$ (Sodium chloride) is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,making it neutral.
$4$. $NH_4Cl$ (Ammonium chloride) is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$. The hydrolysis of $NH_4^+$ ions releases $H^+$ ions,making the solution acidic.

(A) $CuCl_2$ is a salt of a strong acid $(HCl)$ and a weak base $(Cu(OH)_2)$,so its aqueous solution is acidic with $pH < 7$.
$CH_3COONH_4$ is a salt of a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$,so its aqueous solution is approximately neutral with $pH \approx 7$.
$Na_2CO_3$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$,so its aqueous solution is basic with $pH > 7$.
$KNO_3$ is a salt of a strong acid $(HNO_3)$ and a strong base $(KOH)$,so its aqueous solution is neutral with $pH = 7$.

(C) Salts derived from a weak acid or a weak base undergo hydrolysis in aqueous solution.
$NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,so it undergoes cationic hydrolysis.
$Na_2SO_4$ is a salt of a strong acid $(H_2SO_4)$ and a strong base $(NaOH)$,so it does not undergo hydrolysis.
$KCl$ is a salt of a strong acid $(HCl)$ and a strong base $(KOH)$,so it does not undergo hydrolysis.
$KNO_3$ is a salt of a strong acid $(HNO_3)$ and a strong base $(KOH)$,so it does not undergo hydrolysis.

(B) An acidic solution turns blue litmus paper red.
* $NH_4F$ is a salt of a weak acid $HF$ $(K_a = 7.2 \times 10^{-4})$ and a weak base $NH_4OH$ $(K_b = 1.8 \times 10^{-5})$.
Since $K_a > K_b$,the aqueous solution of $NH_4F$ is acidic in nature.
* $NH_4CN$ is a salt of a weak acid $(HCN)$ and a weak base $(NH_4OH)$,where $K_b > K_a$,making it basic.
* $CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$,making it basic.
* $CH_3COONH_4$ is a salt of a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$ with $K_a \approx K_b$,making it neutral.

(D) The dissociation of $AgCl$ is given by: $AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}$.
For a $1:1$ electrolyte,the solubility product $K_{sp} = s^2$,where $s$ is the solubility.
Given $s = 7.2 \times 10^{-7} \ mol \ dm^{-3}$.
$K_{sp} = (7.2 \times 10^{-7})^2$.
$K_{sp} = 51.84 \times 10^{-14}$.
$K_{sp} = 5.184 \times 10^{-13} \approx 5.18 \times 10^{-13}$.

(A) The dissociation of $AgCl$ is given by: $AgCl_{(s)} \rightleftharpoons Ag^{+}_{(aq)} + Cl^{-}_{(aq)}$
For a $1:1$ electrolyte,the solubility product $K_{sp}$ is related to solubility $S$ by the equation: $K_{sp} = S^2$
Given $S = 1.25 \times 10^{-5} \ mol \ dm^{-3}$
$K_{sp} = (1.25 \times 10^{-5})^2 = 1.5625 \times 10^{-10} \approx 1.56 \times 10^{-10}$

(C) The dissociation of the salt is given by: $AX_{2(s)} \rightleftharpoons A^{2+}_{(aq)} + 2X^{-}_{(aq)}$
Let $S$ be the solubility of the salt.
At equilibrium,$[A^{2+}] = S$ and $[X^{-}] = 2S$.
The solubility product expression is: $K_{sp} = [A^{2+}][X^{-}]^2$
$K_{sp} = (S)(2S)^2 = 4S^3$
Given $K_{sp} = 3.2 \times 10^{-8}$.
$4S^3 = 3.2 \times 10^{-8}$
$S^3 = \frac{3.2}{4} \times 10^{-8} = 0.8 \times 10^{-8} = 8 \times 10^{-9}$
$S = \sqrt[3]{8 \times 10^{-9}} = 2 \times 10^{-3} \ mol \ dm^{-3}$

(D) The dissociation of $AgBr$ is given by: $AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)$.
Let the solubility be $S \ mol \ dm^{-3}$.
Then,$[Ag^+] = S$ and $[Br^-] = S$.
The solubility product expression is $K_{sp} = [Ag^+][Br^-] = S \times S = S^2$.
Given $K_{sp} = 4.9 \times 10^{-13}$.
So,$S^2 = 4.9 \times 10^{-13}$.
$S = \sqrt{4.9 \times 10^{-13}} = \sqrt{49 \times 10^{-14}} = 7.0 \times 10^{-7} \ mol \ dm^{-3}$.

(B) The dissociation of silver oxalate is given by: $Ag_2C_2O_{4(s)} \rightleftharpoons 2Ag^+{(aq)} + C_2O_4^{2-}{(aq)}$
Let the solubility be $S = 2 \times 10^{-4} \ mol \ L^{-1}$.
Then,$[Ag^+] = 2S$ and $[C_2O_4^{2-}] = S$.
The solubility product expression is: $K_{sp} = [Ag^+]^2 [C_2O_4^{2-}]$
Substituting the values: $K_{sp} = (2S)^2 \times (S) = 4S^3$
Calculating the value: $K_{sp} = 4 \times (2 \times 10^{-4})^3 = 4 \times (8 \times 10^{-12}) = 3.2 \times 10^{-11}$

(C) The dissociation of the salt is given by: $AB_{2(s)} \rightleftharpoons A^{2+}_{(aq)} + 2B^-_{(aq)}$
Let the solubility be $S = 1.0 \times 10^{-4} \ mol \ dm^{-3}$.
At equilibrium,the concentration of $[A^{2+}] = S$ and $[B^-] = 2S$.
The solubility product expression is: $K_{sp} = [A^{2+}][B^-]^2$.
Substituting the values: $K_{sp} = (S) \cdot (2S)^2 = 4S^3$.
$K_{sp} = 4 \times (1.0 \times 10^{-4})^3 = 4 \times 10^{-12}$.

(A) The dissociation of calcium phosphate in water is given by the equation: $Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}{(aq)} + 2PO_4^{3-}{(aq)}$
The solubility product constant $(K_{sp})$ is defined as the product of the molar concentrations of the ions,each raised to the power of its stoichiometric coefficient in the balanced chemical equation.
Therefore,$K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2$.

(D) $SO_3$ is an acidic oxide.
$Na_2O$ is a basic oxide.
$N_2O$ is a neutral oxide.
$Al_2O_3$ is an amphoteric oxide,as it reacts with both acids and bases.

(D) Water gas is a fuel gas that consists of a mixture of carbon monoxide $(CO)$ and hydrogen $(H_2)$.
It is produced by passing steam over red-hot coke:
$C_{(s)} + H_2O_{(g)} \xrightarrow{1273 \ K} CO_{(g)} + H_{2_{(g)}}$
Therefore,the correct composition is $CO_{(g)} + H_{2_{(g)}}$.

(C) The oxidation states of the elements in the reaction are: $\stackrel{+3}{C_2}O_4^{2-} + \stackrel{+7}{Mn}O_4^{-} + H^{+}$ $\longrightarrow \stackrel{+2}{Mn^{2+}} + \stackrel{+4}{C}O_2 + H_2O$.
In this reaction,the oxidation number of carbon $(C)$ increases from $+3$ in $C_2O_4^{2-}$ to $+4$ in $CO_2$.
Since the oxidation number increases,$C_2O_4^{2-}$ undergoes oxidation and acts as the reducing agent (reductant).

(A) To identify the reductant,we determine the oxidation states of the elements in the reaction:
$H_2S^{-2} + NO_2^{+4} \rightarrow H_2O + NO^{+2} + S^0$
In $H_2S$,the oxidation state of $S$ increases from $-2$ to $0$. This is an oxidation process.
Since $H_2S$ undergoes oxidation,it acts as the reducing agent (reductant).
In $NO_2$,the oxidation state of $N$ decreases from $+4$ to $+2$. This is a reduction process,so $NO_2$ acts as the oxidizing agent.
Therefore,$H_2S$ is the reductant.

(B) To identify the reducing agent,we analyze the change in oxidation states:
$1$. In $H_2O_{2(aq)}$,the oxidation state of oxygen is $-1$. In $O_{2(g)}$,the oxidation state of oxygen is $0$.
$2$. Since the oxidation state of oxygen increases from $-1$ to $0$,$H_2O_{2(aq)}$ undergoes oxidation.
$3$. $A$ substance that undergoes oxidation acts as a reducing agent.
$4$. Therefore,$H_2O_{2(aq)}$ is the reducing agent.

(C) redox reaction involves a change in the oxidation state of elements.
In the reaction $H_2SO_4 + Ca(OH)_2 \longrightarrow CaSO_4 + 2H_2O$,the oxidation states of all elements $(H, S, O, Ca)$ remain unchanged.
Specifically,$H$ is $+1$,$S$ is $+6$,$O$ is $-2$,and $Ca$ is $+2$ on both sides of the equation.
This is an acid-base neutralization reaction,which is a type of double-displacement reaction,not a redox reaction.

(B) The chemical formula for potassium dichromate is $K_2Cr_2O_7$.
Let the oxidation state of $Cr$ be $x$.
The sum of oxidation states of all atoms in a neutral molecule is $0$.
$2(+1) + 2(x) + 7(-2) = 0$
$2 + 2x - 14 = 0$
$2x - 12 = 0$
$2x = 12$
$x = +6$

(A) The ion $I_3^{-}$ is composed of three iodine atoms.
Let the average oxidation state of each iodine atom be $x$.
The sum of the oxidation states of all atoms in the ion must equal the charge of the ion.
Therefore,$3x = -1$.
Solving for $x$,we get $x = -(1/3)$.

(C) For $NO_3^{-}$: Let the oxidation number of $N$ be $x$.
$x + 3(-2) = -1$
$x - 6 = -1$
$x = +5$
For $NO_2$: Let the oxidation number of $N$ be $y$.
$y + 2(-2) = 0$
$y - 4 = 0$
$y = +4$
Thus,the oxidation number of nitrogen changes from $+5$ to $+4$.

(D) In Stock notation,the oxidation state of the central metal atom is represented by Roman numerals in parentheses.
For $SnCl_4$,the oxidation state of $Sn$ is $+4$,so it is correctly represented as $Sn(IV)Cl_4$.
For $FeCl_3$,the oxidation state of $Fe$ is $+3$,so it should be $Fe(III)Cl_3$.
For $MnO_2$,the oxidation state of $Mn$ is $+4$,so it should be $Mn(IV)O_2$.
For $AuCl_3$,the oxidation state of $Au$ is $+3$,so it should be $Au(III)Cl_3$.

(D) The given reaction is: $x H_2O_2 + ClO_4^{-} \rightarrow x O_2 + ClO_2^{-} + 2 H_2O$
Step $1$: Write the half-reactions.
Reduction half-reaction: $ClO_4^{-} \rightarrow ClO_2^{-}$
Balancing oxygen and charge: $ClO_4^{-} + 4 H^{+} + 4 e^{-} \rightarrow ClO_2^{-} + 2 H_2O$
Step $2$: Oxidation half-reaction.
$H_2O_2 \rightarrow O_2 + 2 H^{+} + 2 e^{-}$
Step $3$: Equalize the number of electrons.
Multiply the oxidation half-reaction by $2$ to match the $4 e^{-}$ in the reduction half-reaction:
$2 H_2O_2 \rightarrow 2 O_2 + 4 H^{+} + 4 e^{-}$
Step $4$: Add the two half-reactions.
$(ClO_4^{-} + 4 H^{+} + 4 e^{-}) + (2 H_2O_2) \rightarrow (ClO_2^{-} + 2 H_2O) + (2 O_2 + 4 H^{+} + 4 e^{-})$
Simplifying,we get:
$2 H_2O_2 + ClO_4^{-} \rightarrow 2 O_2 + ClO_2^{-} + 2 H_2O$
Comparing this with the given equation $x H_2O_2 + ClO_4^{-} \rightarrow x O_2 + ClO_2^{-} + 2 H_2O$,we find $x = 2$.

(D) The chemical formula of crystalline lithium chloride is $LiCl \cdot 2H_2O$.
Therefore,there are $2$ water molecules present in its crystalline structure.

(A) Alkali metals have a general electronic configuration of $ns^1$. When they form unipositive ions $(M^+)$,they lose their only valence electron,resulting in a stable noble gas configuration with no unpaired electrons. Therefore,their compounds are diamagnetic,not paramagnetic.
Alkali metals are indeed soft and can be cut with a knife.
They possess low density; for instance,$Li$,$Na$,and $K$ float on water.
Due to their low ionization energies,they readily lose their valence electron,making them the most electropositive elements.

(B) Alkali metals have lower ionization energies compared to alkaline earth metals,making them more electropositive.
Alkaline earth metals have two valence electrons and form $M^{2+}$ ions with a stable inert gas configuration.
Most compounds of alkaline earth metals are diamagnetic and colourless because they do not have unpaired electrons.
Therefore,the statement that alkaline earth metals are more electropositive than alkali metals is incorrect.

(D) Beryllium $(Be)$ is the only alkaline earth metal that does not react with water or steam even at high temperatures.
This is due to its very small atomic size and high ionization enthalpy compared to other elements in the group.

(A) Lithium $(Li)$ shows a diagonal relationship with Magnesium $(Mg)$.
Due to its small ionic size and high charge density,lithium differs from other alkali metals but resembles magnesium,as the ionic size of $Li^+$ is very close to that of $Mg^{2+}$.

(C) The statement "Mass can neither be created nor destroyed" is the fundamental definition of the $Law \ of \ conservation \ of \ mass$.
This law was proposed by $Antoine \ Lavoisier$ in $1789$.
It states that in any chemical reaction, the total mass of the reactants is equal to the total mass of the products.

(B) The molar mass of ethane $(C_2H_6)$ is $(2 \times 12) + (6 \times 1) = 30 \ g/mol$.
Number of moles of ethane $= \frac{\text{mass}}{\text{molar mass}} = \frac{75 \ g}{30 \ g/mol} = 2.5 \ mol$.
At $S$.$T$.$P$.,$1 \ mol$ of any ideal gas occupies $22.4 \ dm^3$.
Volume occupied $= 2.5 \ mol \times 22.4 \ dm^3/mol = 56 \ dm^3$.

(B) The molar mass of methane $(CH_4)$ is $12 + (4 \times 1) = 16 \ g/mol$.
Number of moles of methane $(CH_4) = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{16 \ g}{16 \ g/mol} = 1 \ mol$.
At $STP$,the volume occupied by $1 \ mol$ of any ideal gas is $22.4 \ L$.
Since $1 \ L = 1000 \ cm^3$,the volume is $22.4 \times 1000 \ cm^3 = 22400 \ cm^3$.

(B) The number of moles is calculated using the formula:
$n = \frac{\text{Given mass}}{\text{Molar mass}}$
Given mass $= 5.4 \ g$
Molar mass $= 60 \ g/mol$
$n = \frac{5.4}{60} = 0.09 \ mol$

(B) At $STP$,the molar volume of an ideal gas is $22.4 \ L$ or $22400 \ cm^3$.
Number of moles of gas $= \frac{\text{Volume at STP}}{22400 \ cm^3/mol} = \frac{22400 \ cm^3}{22400 \ cm^3/mol} = 1 \ mol$.
Number of molecules $= \text{moles} \times N_A = 1 \times 6.022 \times 10^{23} = 6.022 \times 10^{23}$ molecules.

(D) At $S.T.P.$,the molar volume of an ideal gas is $22.4 \ dm^3 \ mol^{-1}$.
Number of moles of methane $(CH_4) = \frac{\text{Given volume}}{\text{Molar volume at } S.T.P.} = \frac{33.6 \ dm^3}{22.4 \ dm^3 \ mol^{-1}} = 1.5 \ mol$.
The molar mass of methane $(CH_4) = 12 + (4 \times 1) = 16 \ g \ mol^{-1}$.
Mass of methane $= \text{moles} \times \text{molar mass} = 1.5 \ mol \times 16 \ g \ mol^{-1} = 24 \ g$.
Converting to $kg$: $24 \ g = 24 \times 10^{-3} \ kg = 2.4 \times 10^{-2} \ kg$.

(C) $\text{Number of moles of gas} = \frac{\text{Given volume}}{22.4 \ dm^3/mol}$
$\text{Number of moles} = \frac{67.2}{22.4} = 3 \ mol$
$\text{Number of molecules} = \text{moles} \times N_A$
$\text{Number of molecules} = 3 \times 6.022 \times 10^{23} \approx 3 \times 6 \times 10^{23}$
$\text{Number of molecules} = 18 \times 10^{23} = 1.8 \times 10^{24}$

(C) The formula for calculating mass from moles is: $\text{Mass} = \text{moles} \times \text{Molar mass}$.
The molar mass of $H_2O$ is $(2 \times 1.008) + 16.00 = 18.016 \ g/mol \approx 18 \ g/mol$.
Given moles $= 0.25 \ mol$.
Therefore,$\text{Mass} = 0.25 \ mol \times 18 \ g/mol = 4.5 \ g$.

(A) The balanced chemical equation for the reaction is: $H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$
From the stoichiometry,$1 \ \text{mole}$ of $H_2O$ $(18 \ g)$ is produced by $0.5 \ \text{moles}$ of $O_2$.
At $S.T.P$,$1 \ \text{mole}$ of gas occupies $22.4 \ dm^3$,so $0.5 \ \text{moles}$ of $O_2$ occupies $11.2 \ dm^3$.
Thus,$11.2 \ dm^3$ of $O_2$ is required to produce $18 \ g$ of $H_2O$.
For $9 \ g$ of $H_2O$,the volume of $O_2$ required is: $\frac{11.2 \ dm^3}{18 \ g} \times 9 \ g = 5.6 \ dm^3$.

(C) The balanced chemical equation for the decomposition of potassium chlorate is:
$2KClO_3 \longrightarrow 2KCl + 3O_2$
Calculate the molar mass of $KClO_3$:
$M(KClO_3) = 39 + 35.5 + 3 \times 16 = 122.5 \ g/mol$
Calculate the molar mass of $KCl$:
$M(KCl) = 39 + 35.5 = 74.5 \ g/mol$
Calculate the moles of $KClO_3$ present:
$n(KClO_3) = \frac{12.25 \ g}{122.5 \ g/mol} = 0.1 \ mol$
From the stoichiometry of the reaction,$2 \ mol$ of $KClO_3$ produces $2 \ mol$ of $KCl$. Therefore,$0.1 \ mol$ of $KClO_3$ will produce $0.1 \ mol$ of $KCl$.
Calculate the mass of $KCl$ produced:
$Mass = n \times M = 0.1 \ mol \times 74.5 \ g/mol = 7.45 \ g$

(B) $\text{Percentage atom economy} = \frac{\text{Formula weight of the desired product}}{\text{Sum of formula weight of all the reactants}} \times 100$
$\text{Percentage atom economy} = \frac{46}{92} \times 100 = 50 \%$

(C) Atom economy is a measure of the efficiency of a chemical reaction in terms of how many atoms from the reactants end up in the desired final product.
It is calculated using the formula:
$\%$ Atom economy $= \frac{\text{Formula weight of desired product}}{\text{Sum of formula weight of all reactants}} \times 100$

(B) Atom economy is the conversion efficiency of a chemical process in terms of all atoms involved and the desired products produced.
$\text{Percentage Atom Economy} = \frac{\text{Formula weight of desired product}}{\text{Sum of formula weight of all reactants}} \times 100$
$= \frac{123}{246} \times 100$
$= 50.00 \%$

(D) According to Boyle's Law,at constant temperature,$P_1 V_1 = P_2 V_2$.
Given: $P_1 = 105 \ kPa$,$V_1 = 11.2 \ dm^3$,$P_2 = 420 \ kPa$.
Substituting the values: $105 \times 11.2 = 420 \times V_2$.
$V_2 = \frac{105 \times 11.2}{420}$.
$V_2 = 2.8 \ dm^3$.

(D) The correct answer is $(d)$.
Transition elements (also known as transition metals) are elements that have partially filled $d$ orbitals.
$IUPAC$ defines transition elements as an element having a $d$ subshell that is partially filled with electrons,or an element that has the ability to form stable cations with an incompletely filled $d$ orbital.
In general,any element which corresponds to the $d$-block of the modern periodic table (which consists of groups $3-12$) is considered to be a transition element.

(D) . Besides $EO_2$ type,$S$,$Se$,and $Te$ also form $EO_3$ type oxides.
Group $16$ elements consist of non-metals ($O$,$S$),metalloids ($Se$,$Te$),and a metal $(Po)$.
All elements of this group require $2$ electrons to attain a stable noble gas configuration.
The reducing character of dioxides decreases from $SO_2$ to $TeO_2$ because the stability of the $+4$ oxidation state increases down the group due to the inert pair effect.

(B) neutral complex is a coordination compound that carries no net charge on the coordination sphere.
In the complex $[Co(NO_2)_3(NH_3)_3]$,the oxidation state of $Co$ is $+3$,the charge of three $NO_2^-$ ligands is $-3$,and $NH_3$ is a neutral ligand.
Therefore,the net charge is $(+3) + (-3) + 0 = 0$.
Thus,$[Co(NO_2)_3(NH_3)_3]$ is a neutral complex.

(A) The complex $[Pt(NH_3)_2 Cl_2]$ is a neutral coordination compound because it carries no net charge.
In this complex,both $NH_3$ and $Cl^-$ ligands are directly bonded to the central metal atom $Pt$ within the coordination sphere.
The coordination number of $Pt$ is determined by the number of donor atoms attached to it,which is $2 (NH_3) + 2 (Cl) = 4$.
The oxidation state of $Pt$ is calculated as: $x + 2(0) + 2(-1) = 0$,which gives $x = +2$.

(C) According to the spectrochemical series,the order of ligand field strength is:
$S^{2-} < H_2O < EDTA^{4-} < en$.
Therefore,$en$ (ethylenediamine) has the highest field strength among the given options.

(A) neutral ligand is a ligand that carries no net electrical charge.
$1$. Thiocyanate $[SCN]^-$ is an anionic ligand with a charge of $-1$.
$2$. Ammine $[NH_3]$ is a neutral ligand.
$3$. Aqua $[H_2O]$ is a neutral ligand.
$4$. Carbon monoxide $[CO]$ is a neutral ligand.
Therefore,Thiocyanate is not a neutral ligand.

(A) Complex compounds in which all the ligands bound to the metal center are identical are known as homoleptic complexes.
Complexes in which the metal atom or ion is linked with more than one kind of ligand are known as heteroleptic complexes.
$1. \left[Co(NH_3)_6\right] Cl_3$: All ligands are $NH_3$ (identical),so it is a homoleptic complex.
$2. \left[Co(NH_3)_4 Cl_2\right]$: Ligands are $NH_3$ and $Cl^-$ (different),so it is a heteroleptic complex.
$3. \left[Co(NH_3)_5 Cl\right] SO_4$: Ligands are $NH_3$ and $Cl^-$ (different),so it is a heteroleptic complex.
$4. \left[Co(ONO)(NH_3)_5\right] Cl_2$: Ligands are $ONO^-$ and $NH_3$ (different),so it is a heteroleptic complex.

(D) cationic complex is a coordination compound in which the complex ion carries a net positive charge.
In the option $D$,the complex $[PtBr_2(NH_3)_4] Br_2$ dissociates as:
$[PtBr_2(NH_3)_4] Br_2 \rightarrow [PtBr_2(NH_3)_4]^{2+} + 2Br^{-}$
Here,the complex ion $[PtBr_2(NH_3)_4]^{2+}$ has a positive charge,making it a cationic complex.
Options $A$ and $C$ are anionic complexes,and option $B$ is a neutral complex.

(D) The effective atomic number $(EAN)$ is the total number of electrons present with the central metal atom or ion in a coordination complex,including the electrons donated by the ligands.
The formula to calculate the $EAN$ is:
$EAN = Z - O.N. + 2 \times C.N.$
Where:
$Z = \text{Atomic number} = 27$
$O.N. = \text{Oxidation number of } Co \text{ in } [Co(NH_3)_6]^{3+} = +3$
$C.N. = \text{Coordination number of } Co = 6$
Substituting the values:
$EAN = 27 - 3 + 2 \times 6 = 24 + 12 = 36$.

(A) Ethylene diaminetetraacetic acid $(EDTA)$ is an amino polycarboxylic acid with the formula $(HO_2CCH_2)_2NCH_2CH_2N(CH_2CO_2H)_2$.
In this structure,there are two nitrogen atoms present in the ethylene diamine backbone.
Therefore,the number of $N$ atoms in $EDTA$ is $2$.

(B) For the complex $[Ni(CN)_4]^{2-}$:
$1$. Let the oxidation state of $Ni$ be $x$.
$x + 4(-1) = -2 \Rightarrow x = +2$.
Thus,the oxidation state of $Ni$ is $+2$,not $+6$.
$2$. $CN^-$ is a strong field ligand,which causes the pairing of electrons in the $3d$ orbitals.
$3$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
$4$. Due to the pairing,the complex undergoes $dsp^2$ hybridization,resulting in a square planar geometry.

(A) Complex compounds in which all the ligands bound to the metal center are identical are known as homoleptic complexes.
Complexes in which the metal atom or ion is linked with more than one kind of ligand are known as heteroleptic complexes.
In $[Co(NH_3)_6]^{3+}$,all six ligands are $NH_3$ molecules,which are identical. Therefore,it is a homoleptic complex.
In the other options,there are different types of ligands (e.g.,$NH_3$ and $Cl^-$,or $NH_3$ and $H_2O$),making them heteroleptic complexes.

(A) The complex is $Na_4[Fe(CN)_6]$.
$1$. The complex ion is $[Fe(CN)_6]^{4-}$,which carries a $-4$ charge.
$2$. To find the oxidation state of $Fe$ $(x)$:
$x + 6(-1) = -4$
$x - 6 = -4$
$x = +2$
$3$. The coordination number $(C.N.)$ of $Fe$ is determined by the number of ligands attached,which is $6$ ($CN^-$ is a monodentate ligand).
Therefore,the statement that the complex ion carries a $-4$ charge is true.

(B) The effective atomic number $(EAN)$ is calculated using the formula: $EAN = Z - X + Y$,where:
$Z$ = atomic number of the metal = $78$
$X$ = number of electrons lost by the metal to form the ion = $2$ (since the oxidation state of $Pt$ in $[Pt(NH_3)_4]^{2+}$ is $+2$)
$Y$ = number of electrons donated by the ligands = $4 \times 2 = 8$ (since there are $4$ $NH_3$ ligands,each donating $2$ electrons)
$EAN = 78 - 2 + 8 = 84$

(D) The Effective Atomic Number $(EAN)$ is calculated using the formula: $EAN = Z - x + Y$,where $Z$ is the atomic number,$x$ is the oxidation state,and $Y$ is the number of electrons donated by ligands.
Given: $EAN = 36$,$Z = 27$,and $C.N. = 6$.
Since each ligand donates $2$ electrons,$Y = 6 \times 2 = 12$.
Substituting the values into the formula:
$36 = 27 - x + 12$
$36 = 39 - x$
$x = 39 - 36$
$x = +3$
Therefore,the oxidation state of cobalt is $+3$.

(D) The spectrochemical series arranges ligands in the order of their increasing field strength:
$I^{-} < Br^{-} < S^{2-} < Cl^{-} < OH^{-} < C_2O_4^{2-} < H_2O < NCS^{-} < Py < NH_3 < en < NO_2^{-} < CN^{-} < CO$
Comparing the given ligands,the correct order of increasing field strength is:
$I^{-} < S^{2-} < OH^{-} < NH_3$

(A) The given complex is $[Co(H_2O)(NH_3)_5]I_3$.
First,determine the oxidation state of the central metal atom $Co$:
Let the oxidation state of $Co$ be $x$.
$x + 1(0) + 5(0) + 3(-1) = 0$
$x - 3 = 0$
$x = +3$
According to $IUPAC$ nomenclature rules for coordination compounds:
$1$. Name the ligands in alphabetical order: 'ammine' $(NH_3)$ comes before 'aqua' $(H_2O)$.
$2$. Since there are $5$ ammine ligands,it is 'pentaammine'.
$3$. Since there is $1$ aqua ligand,it is 'aqua'.
$4$. The metal is in a cationic complex,so it is named 'cobalt'.
$5$. The oxidation state is written in Roman numerals in parentheses: $(III)$.
$6$. The counter ion is 'iodide'.
Combining these,the name is 'Pentaammineaquacobalt$(III)$ iodide'.

(B) In $[NiCl_4]^{2-}$,$Ni$ is in $+2$ oxidation state: $Ni^{2+} \rightarrow [Ar] 3d^8$.
Since $Cl^-$ is a weak field ligand,it does not cause pairing of $3d$ electrons.
Thus,the hybridization involved is $sp^3$ using $4s$ and $4p$ orbitals.
In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state: $Ni^{2+} \rightarrow [Ar] 3d^8$.
$CN^-$ is a strong field ligand,which causes pairing of $3d$ electrons.
This leaves one $3d$ orbital vacant,resulting in $dsp^2$ hybridization using one $3d$,one $4s$,and two $4p$ orbitals.

(C) The stability of a metal complex is inversely proportional to the size of the metal cation.
As the size of the cation increases,the electrostatic attraction between the metal ion and the ligand decreases,leading to lower stability.
Among the given cations ($Cu^{2+}$,$Fe^{2+}$,$Cd^{2+}$,$Ni^{2+}$),$Cd^{2+}$ has the largest ionic radius.
Therefore,$Cd^{2+}$ forms the complex with the lowest stability.

(D) For $[CoF_6]^{3-}$,$Co^{3+}$ is $3d^6$. It undergoes $sp^3d^2$ hybridization,resulting in an octahedral geometry and it is paramagnetic.
For $[Co(NH_3)_6]^{3+}$,$Co^{3+}$ is $3d^6$. It undergoes $d^2sp^3$ hybridization due to the strong field ligand $NH_3$,resulting in an octahedral geometry and it is diamagnetic.
For $[NiCl_4]^{2-}$,$Ni^{2+}$ is $3d^8$. It undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry and it is paramagnetic.
For $[Ni(CN)_4]^{2-}$,$Ni^{2+}$ is $3d^8$. The strong field ligand $CN^-$ causes pairing of $3d$ electrons,leading to $dsp^2$ hybridization,a square planar geometry,and it is diamagnetic.

(D) For the complex $[Co(NH_3)_6]^{3+}$:
$1$. The oxidation state of $Co$ is $+3$.
$2$. The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6 4s^0$.
$3$. $NH_3$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.
$4$. After pairing,all $6$ electrons occupy the first three $3d$ orbitals,leaving no unpaired electrons $(n = 0)$.
$5$. The complex undergoes $d^2sp^3$ hybridization,resulting in an octahedral geometry.

(B) The electronic configuration of $Cd$ $(Z=48)$ is $[Kr] 4d^{10} 5s^2$.
Since the $4d$ subshell contains $10$ electrons,it is completely filled.

(D) Interstitial compounds are formed when small atoms like $H, C, N$ are trapped inside the crystal lattice of transition metals.
They possess the following properties:
$1$. They have high melting points,higher than those of pure metals.
$2$. They are very hard.
$3$. They retain metallic conductivity.
$4$. They are chemically inert.
Therefore,the statement that metallic carbides are chemically inert is correct.

(C) Interstitial compounds are formed when small atoms like $H$,$C$,or $N$ are trapped inside the crystal lattice of transition metals.
- These compounds retain the chemical properties of the parent transition metals,meaning they are chemically inert.
- They possess very high melting points,which are higher than those of the pure parent metals.
- They are extremely hard,with some borides being comparable to diamond in hardness.
- They maintain the electrical and thermal conductivity of the parent metal.
- Therefore,the statement that their chemical properties are different from the parent metal is incorrect.

(A) The electronic configuration of Manganese $(Mn)$ is $[Ar] 3d^5 4s^2$.
Due to the presence of $7$ valence electrons ($5$ in the $3d$ subshell and $2$ in the $4s$ subshell),it can exhibit a wide range of oxidation states from $+2$ to $+7$.

(A) The electronic configuration of lanthanoids follows the general pattern $(Xe)4f^{n} 5d^{0-1} 6s^{2}$.
$A$ half-filled $f$-orbital corresponds to $f^{7}$.
Looking at the provided table:
- $Ce$ $(Z=58)$: $(Xe)4f^{1} 5d^{1} 6s^{2}$
- $Pm$ $(Z=61)$: $(Xe)4f^{5} 6s^{2}$
- $Sm$ $(Z=62)$: $(Xe)4f^{6} 6s^{2}$
- $Eu$ $(Z=63)$: $(Xe)4f^{7} 6s^{2}$
Since $Eu$ has $4f^{7}$,it possesses a half-filled $f$-orbital. This configuration is stable and matches the expected filling order for this element.

(C) Promethium is a chemical element with the symbol $Pm$ and atomic number $61$.
All of its isotopes are radioactive.
It is extremely rare,with only about $500-600 \ g$ naturally occurring in Earth's crust at any given time.

(D) Nichrome is an alloy of $Ni$ and $Cr$. Due to its high resistance to oxidation and heat,it is used in heating elements and gas turbine engines.

(B) In $Zn$ $(3d^{10} 4s^2)$,all electrons in the $d$-orbitals are paired.
Due to the absence of unpaired electrons,the metallic bonds in $Zn$ are weak.
Consequently,$Zn$ is a soft metal compared to other transition elements.

(A) Brass is an alloy of copper $(Cu)$ and zinc $(Zn)$.
It does not contain iron $(Fe)$,therefore it is classified as a nonferrous alloy.

(C) The colour of transition metal ions depends on the presence of unpaired electrons in the $d$-orbitals,which allow for $d-d$ transitions.
$Cu^{+}$ has an electronic configuration of $[Ar] 3d^{10}$.
Since all $10$ electrons are paired in the $3d$ subshell,there are no unpaired electrons.
Therefore,$Cu^{+}$ does not exhibit $d-d$ transitions and does not form coloured compounds.

(A) The electronic configuration of $Sc$ is $[Ar] 3d^1 4s^2$.
In the $+3$ oxidation state,$Sc^{3+}$ has the configuration $[Ar] 3d^0$.
Since $Sc^{3+}$ has no unpaired electrons in its $d$-orbitals,it cannot undergo $d-d$ transitions,making its compounds colourless.

(C) Zinc oxide $(ZnO)$ and Titanium dioxide $(TiO_2)$ nanoparticles are commonly used in sunscreen lotions.
These compounds protect the skin against harmful $UV$ (ultraviolet) rays by absorbing or reflecting the radiation,thereby preventing skin damage.

(A) Lanthanoids exhibit a common oxidation state of $+3$.
When reacting with nitrogen $(N^{3-})$,the compound formed is $LnN$.
When reacting with halogens $(X^-)$,the compound formed is $LnX_3$.

(C) Lanthanoids are metallic in nature and are good conductors of heat and electricity.
Most lanthanoid ions are paramagnetic due to the presence of unpaired $4f$ electrons.
Lanthanoids generally exhibit high coordination numbers,typically greater than $6$ (often $8$ or $9$).
However,the statement that 'all lanthanoids are non-radioactive' is incorrect.
Promethium ($Pm$,atomic number $61$) is a well-known radioactive element among the lanthanoids.

(C) The reduction reaction for calcium is: $Ca^{2+} + 2e^- \longrightarrow Ca_{(s)}$
From the stoichiometry,$1 \ mol$ of $Ca$ requires $2 \ mol$ of electrons,which is equivalent to $2 \ F$ of electricity.
Moles of $Ca$ produced $= \frac{\text{mass}}{\text{molar mass}} = \frac{10 \ g}{40 \ g \ mol^{-1}} = 0.25 \ mol$.
Electricity required $= \text{moles of } Ca \times 2 \ F \ mol^{-1} = 0.25 \times 2 = 0.5 \ F$.

(D) During electrolysis,the reduction reaction occurs at the cathode,where metal ions gain electrons to form metal atoms. Therefore,the metal is deposited at the cathode,not the anode. Thus,option $D$ is incorrect.

(C) The total charge $q$ passed is given by the formula $q = I \times t$.
Given $I = 1 \ A$ and $t = 16.1 \ \text{minutes} = 16.1 \times 60 \ \text{seconds} = 966 \ \text{seconds}$.
Thus,$q = 1 \times 966 = 966 \ \text{Coulombs}$.
The number of electrons $n$ is calculated using $q = n \times e$,where $e = 1.602 \times 10^{-19} \ \text{C}$.
$n = \frac{q}{e} = \frac{966}{1.602 \times 10^{-19}} \approx 6.022 \times 10^{21}$ electrons.

(A) The total charge $Q$ passed is given by the formula $Q = I \times t$.
Given $I = 5 \ A$ and $t = 20 \ minutes = 20 \times 60 \ s = 1200 \ s$.
$Q = 5 \times 1200 = 6000 \ C$.
Since the charge of $1 \ mole$ of electrons is $F = 96500 \ C/mol$,the number of moles of electrons is $n = \frac{Q}{F}$.
$n = \frac{6000}{96500} \approx 0.06217 \ mol \approx 6.22 \times 10^{-2} \ mol$.
Rounding to the nearest provided option,the value is $6.25 \times 10^{-2}$.

(B) The reduction reaction at the cathode is: $Ca^{2+} + 2e^- \longrightarrow Ca_{(s)}$.
From the reaction,the $n$-factor $(nf)$ for calcium is $2$.
The molar mass of calcium is $40 \ g/mol$.
The number of moles of calcium deposited is: $\text{moles} = \frac{0.8 \ g}{40 \ g/mol} = 0.02 \ mol$.
The number of equivalents of calcium is: $\text{equivalents} = \text{moles} \times nf = 0.02 \times 2 = 0.04 \ eq$.
According to Faraday's law of electrolysis,$1 \ Faraday$ of electricity is required to deposit $1 \ equivalent$ of a substance.
Therefore,to deposit $0.04 \ equivalents$ of calcium,$0.04 \ F$ of electricity is required.

(D) The relationship between standard Gibbs energy change $(\Delta G^{\circ})$ and standard cell potential $(E_{\text{cell}}^{\circ})$ is given by the equation:
$\Delta G^{\circ} = -nFE_{\text{cell}}^{\circ}$
Where $n$ is the number of moles of electrons transferred,$F$ is the Faraday constant,and $E_{\text{cell}}^{\circ}$ is the standard cell potential.

(B) The total charge $Q$ flowing through the wire is given by $Q = I \times t$.
Given $I = 1.5 \ A$ and $t = 3 \ hours = 3 \times 3600 \ s = 10800 \ s$.
So,$Q = 1.5 \times 10800 = 16200 \ C$.
The number of electrons $n$ is given by $n = \frac{Q}{e}$,where $e = 1.6 \times 10^{-19} \ C$.
$n = \frac{16200}{1.6 \times 10^{-19}} = 1.0125 \times 10^{23} \approx 1.01 \times 10^{23}$.

(C) In an electrochemical cell,the species that undergoes oxidation acts as the reducing agent.
At the anode,oxidation occurs: $Ni_{(s)} \rightarrow Ni^{2+} + 2e^{-}$.
Since $Ni_{(s)}$ loses electrons,it is the reducing agent.
At the cathode,reduction occurs: $Ag^{+} + e^{-} \rightarrow Ag_{(s)}$.

(D) According to Faraday's first law of electrolysis,the mass of substance deposited is given by $w = \frac{Z \cdot I \cdot t}{96500}$,where $Z = \frac{\text{Equivalent mass}}{1} = \frac{27}{3} = 9 \ g/mol$.
Given $I = 1 \ A$,$t = 9650 \ s$,and $F = 96500 \ C/mol$.
$w = \frac{9 \times 1 \times 9650}{96500} = \frac{9 \times 9650}{96500} = 0.9 \ g$.
Therefore,the weight of $Al$ deposited is $0.9 \ g$.

(D) $Cu^{2+} + 2e^{-} \longrightarrow Cu$
$1 \ mol \ Cu^{2+} \text{ requires } 2 \ mol \ e^{-}$
$2 \ mol \ Cu^{2+} \text{ requires } 4 \ mol \ e^{-}$
$\text{Charge of } 1 \ mol \text{ electrons } = 96500 \ C$
$\text{Charge required for the reduction of two moles of } Cu^{2+} = 4 \times 96500 = 3.86 \times 10^5 \ C$

(A) In a Galvanic cell,the anode is the negative electrode where oxidation occurs,and the cathode is the positive electrode where reduction occurs.
In a Galvanic cell,chemical energy is converted into electrical energy.

(B) Anode: The electrode of an electrochemical cell at which oxidation occurs.
Cathode: The electrode of an electrochemical cell at which reduction occurs.
Electrolysis of molten $NaCl$ or fused $NaCl$:
Sodium ions $(Na^+)$ migrate to the cathode,where they gain electrons and are reduced to sodium metal:
$Na^+ + e^- \longrightarrow Na_{(s)}$
Chlorine ions $(Cl^-)$ migrate to the anode. They give up their electrons to the anode and are oxidized to chlorine gas:
$Cl^- \longrightarrow \frac{1}{2} Cl_{2(g)} + e^-$
Overall reaction:
$2 NaCl_{(l)} \longrightarrow 2 Na_{(s)} + Cl_{2(g)}$
Therefore,the product obtained at the anode is $Cl_{2(g)}$.

(A) Given: Molarity $(M) = 0.05 \ M$,Conductivity $(\kappa) = 0.0118 \ S \ cm^{-1}$.
The formula for molar conductivity $(\Lambda_{m})$ is:
$\Lambda_{m} = \frac{\kappa \times 1000}{M}$
Substituting the values:
$\Lambda_{m} = \frac{0.0118 \times 1000}{0.05}$
$\Lambda_{m} = \frac{11.8}{0.05} = 236 \ S \ cm^2 \ mol^{-1}$.

(C) The cell constant is determined by measuring the conductivity of a standard aqueous solution of $KCl$.
This is because the conductivity of $KCl$ solutions is known accurately at various concentrations and different temperatures.

(C) Given: Resistance $(R) = 600 \ \Omega$
Conductivity $(\kappa) = 0.0015 \ \Omega^{-1} \ cm^{-1}$
The formula for cell constant $(G^*)$ is:
$G^* = \kappa \times R$
Substituting the values:
$G^* = 0.0015 \ \Omega^{-1} \ cm^{-1} \times 600 \ \Omega = 0.90 \ cm^{-1}$