MHT CET 2020 Chemistry Question Paper with Answer and Solution

$(A)$ We know that $1 \ nm = 10^{-9} \ m$ and $1 \ pm = 10^{-12} \ m$.
To find the relationship between $nm$ and $pm$, we divide the two values:
$\frac{1 \ nm}{1 \ pm} = \frac{10^{-9} \ m}{10^{-12} \ m} = 10^{-9 - (-12)} = 10^{3}$.
Therefore, $1 \ nm = 10^{3} \ pm$.

(A) Given: $n = 2 \ mol$,$T = 546 \ K$,$V = 44.8 \ L$,$R = 0.0821 \ L \ atm \ mol^{-1} \ K^{-1}$.
According to the ideal gas equation,$PV = nRT$.
Therefore,$P = \frac{nRT}{V}$.
Substituting the values: $P = \frac{2 \times 0.0821 \times 546}{44.8}$.
$P = \frac{89.6532}{44.8} \approx 2.0 \ atm$.

(C) Given: $V_{1} = 0.2 \ L$,$T_{1} = 273.15 \ K$.
The final temperature is $T_{2} = 273.15 \ ^{\circ}C = 273.15 + 273.15 = 546.30 \ K$.
According to Charles's Law,at constant pressure $(P)$ and amount of gas $(n)$:
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
Substituting the values:
$V_{2} = \frac{V_{1} \times T_{2}}{T_{1}} = \frac{0.2 \times 546.30}{273.15} = 0.4 \ L$.

(C) Given: $V_{1} = 3.4 \ L$,$T_{1} = 25^{\circ} C = 25 + 273 = 298 \ K$.
$V_{2} = 10.2 \ L$,$T_{2} = ?$.
According to Charles's Law,at constant pressure and amount of gas,$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$.
Substituting the values: $\frac{3.4 \ L}{298 \ K} = \frac{10.2 \ L}{T_{2}}$.
$T_{2} = \frac{10.2 \ L \times 298 \ K}{3.4 \ L} = 3 \times 298 \ K = 894 \ K$.

(D) Given: $V_{1} = 2 \ dm^{3}$,$T_{1} = 273.15 \ K$ (at $STP$).
We need to find $T_{2}$ when $V_{2} = 2 \times V_{1} = 4 \ dm^{3}$ at constant pressure.
According to Charles's law,$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$.
Substituting the values: $\frac{2}{273.15} = \frac{4}{T_{2}}$.
$T_{2} = \frac{4 \times 273.15}{2} = 546.3 \ K$.
To convert to Celsius: $t^{\circ} C = T(K) - 273.15$.
$t^{\circ} C = 546.3 - 273.15 = 273.15^{\circ} C$.

(A) According to Boyle's law,at constant temperature,$P_1 V_1 = P_2 V_2$.
Given: $P_1 = 720 \ mm$,$V_1 = 100 \ mL$,$V_2 = 84 \ mL$.
Substituting the values into the equation: $P_2 = \frac{P_1 V_1}{V_2}$.
$P_2 = \frac{720 \times 100}{84} = \frac{72000}{84} \approx 857.14 \ mm$.

(B) According to Andrews' experiments on carbon dioxide,the gas obeys Boyle's law only at temperatures above its critical temperature.
For carbon dioxide,the critical temperature $(T_c)$ is $31.1^{\circ}C$.
At temperatures below $31.1^{\circ}C$,the gas can be liquefied by pressure,and it deviates significantly from ideal gas behavior.
Therefore,the minimum temperature at which $CO_2$ behaves like an ideal gas (obeying Boyle's law) is its critical temperature,which is $31.1^{\circ}C$.

(A) Given: $n=2 \ mol$,$T=546 \ K$,$V=44.8 \ L$,$R=0.082 \ L \ atm \ K^{-1} \ mol^{-1}$.
According to the ideal gas equation,$PV=nRT$.
Substituting the values: $P = \frac{nRT}{V} = \frac{2 \times 0.082 \times 546}{44.8}$.
$P = \frac{89.544}{44.8} \approx 1.998 \ atm$.
Therefore,the correct option is $A$.

(D) According to Charles's Law,the volume of a fixed mass of gas is directly proportional to its absolute temperature $(V \propto T)$.
As the temperature decreases,the volume of the gas decreases.
At a temperature of $-273.15^{\circ} C$ (which is $0 \ K$),the volume of a gas theoretically becomes zero.
This temperature is known as absolute zero.

(B) The $SI$ unit of viscosity is the Pascal-second $(Pa \ s)$,which is equivalent to $N \ s \ m^{-2}$ or $Kg \ m^{-1} \ s^{-1}$.
Therefore,the correct option is $(B)$.

(B) Given: $P_{1} = 1 \ bar$,$V_{1} = 2.27 \ L$,$P_{2} = 0.227 \ bar$.
According to Boyle's Law,$P_{1}V_{1} = P_{2}V_{2}$ at constant temperature.
Substituting the values: $1 \ bar \times 2.27 \ L = 0.227 \ bar \times V_{2}$.
$V_{2} = \frac{1 \times 2.27}{0.227} = 10 \ L$.
The rise in volume of the gas is $\Delta V = V_{2} - V_{1} = 10 \ L - 2.27 \ L = 7.73 \ L$.

(B) According to Andrews' experiments on $CO_2$ isotherms,the critical pressure $(P_C)$ of carbon dioxide is $73 \ atm$.
At this critical pressure,the temperature at which the gas starts to condense (or the temperature at which the gas becomes indistinguishable from its liquid phase) is known as the critical temperature $(T_C)$.
For $CO_2$,the critical temperature $(T_C)$ is $30.98^{\circ} C$.

(C) Gases with higher critical temperatures and stronger intermolecular forces of attraction are easier to liquefy.
Among the given options,$Cl_{2}$ (chlorine) has a relatively high critical temperature $(417 \ K)$ and strong van der Waals forces,making it easy to liquefy.
In contrast,gases like $O_{2}$,$N_{2}$,and $He$ have very low critical temperatures and are considered permanent gases,which are difficult to liquefy at room temperature.

(D) The molecular weight of water isotopes depends on the mass of the hydrogen isotopes $(H=1, D=2, T=3)$ and oxygen isotopes $(^{16}O=16, ^{17}O=17, ^{18}O=18)$.
Calculating the molar mass for each option:
$A$. $T_{2}{ }^{17}O = (3 \times 2) + 17 = 23 \ g/mol$
$B$. $T_{2}{ }^{16}O = (3 \times 2) + 16 = 22 \ g/mol$
$C$. $HD^{18}O = (1 + 2) + 18 = 21 \ g/mol$
$D$. $T_{2}{ }^{18}O = (3 \times 2) + 18 = 24 \ g/mol$
Comparing the values,$T_{2}{ }^{18}O$ has the highest molecular weight,making it the heaviest isotope of water.

(D) The given electronic configuration is $[Ar] 3d^{10} 4s^{2}$.
Summing the electrons: $18 (Ar) + 10 (3d) + 2 (4s) = 30$.
The atomic number $Z = 30$ corresponds to the element Zinc $(Zn)$.

(D) The electronic configuration of $Cu (Z=29)$ is $[Ar] 3d^{10} 4s^{1}$.
This occurs because a fully filled $3d$ subshell provides extra stability to the atom.
Therefore,$Cu$ has only one electron in the $4s$ orbital.

(D) Caryophyllene is a natural bicyclic sesquiterpene that is a constituent of many essential oils.
It is primarily found in the essential oil of $Syzygium$ $aromaticum$ (cloves),$Cannabis$ $sativa$,rosemary,and hops.
Therefore,the correct source among the given options is oil of cloves.

(D) $I$. Extensive property: Heat capacity (depends on the amount of matter present in the system).
$II$. Intensive property: Surface tension,density,and refractive index (independent of the amount of matter present in the system).

(A) An extensive property is a property whose value depends on the quantity or size of matter present in the system (e.g.,heat,mass,volume).
An intensive property is a property whose value is independent of the quantity or size of matter present in the system (e.g.,temperature,pressure,density).
Therefore,heat is an extensive property and temperature is an intensive property.

(A) Given: $n = 2 \ mol$,$V_{1} = 12.5 \ L = 12.5 \times 10^{-3} \ m^{3}$,$V_{2} = 15.0 \ L = 15.0 \times 10^{-3} \ m^{3}$,$P_{ext} = 760 \ mm \ Hg = 1.013 \times 10^{5} \ N \ m^{-2}$.
Work done in an irreversible expansion is given by the formula: $W = -P_{ext} \Delta V$.
$W = -P_{ext}(V_{2} - V_{1}) = -1.013 \times 10^{5} \times (15.0 - 12.5) \times 10^{-3} \ J$.
$W = -1.013 \times 10^{5} \times 2.5 \times 10^{-3} \ J$.
$W = -1.013 \times 250 \ J$.
$W = -253.25 \ J$.

(D) Given: $V_1 = 10 \ dm^3, V_2 = 100 \ dm^3, T = 300 \ K, R = 8.314 \ J \ K^{-1} \ mol^{-1}$.
Mass of $O_2$ $(m)$ = $1.6 \times 10^{-2} \ kg = 16 \ g$.
Molar mass of $O_2$ $(M)$ = $32 \ g \ mol^{-1}$.
Number of moles $(n)$ = $\frac{m}{M} = \frac{16}{32} = 0.5 \ mol$.
For an isothermal and reversible expansion process,the work done is given by:
$W = -2.303 \ nRT \ \log_{10} \left( \frac{V_2}{V_1} \right)$.
Substituting the values:
$W = -2.303 \times 0.5 \times 8.314 \times 300 \times \log_{10} \left( \frac{100}{10} \right)$.
$W = -2.303 \times 0.5 \times 8.314 \times 300 \times \log_{10} (10)$.
Since $\log_{10} (10) = 1$,
$W = -2.303 \times 0.5 \times 8.314 \times 300 = -2872 \ J$.

(C) Given: $V_{1} = 4 \ dm^{3}$,$V_{2} = 6 \ dm^{3}$,$P_{ex} = 3 \ atm$.
The formula for work done against constant external pressure is $W = -P_{ex} \Delta V$.
$\Delta V = V_{2} - V_{1} = 6 \ dm^{3} - 4 \ dm^{3} = 2 \ dm^{3} = 2 \ L$.
$W = -3 \ atm \times 2 \ L = -6 \ L \cdot atm$.
Since $1 \ L \cdot atm = 101.3 \ J$,we have $W = -6 \times 101.3 \ J = -607.8 \ J$.

(A) The formula for work done during expansion against a constant external pressure is $W = -P_{ex} \Delta V$.
Given:
$V_{1} = 1 \times 10^{-3} \ m^{3} = 0.001 \ m^{3}$
$V_{2} = 1 \times 10^{-2} \ m^{3} = 0.01 \ m^{3}$
$P_{ex} = 1 \times 10^{5} \ Nm^{-2}$
Change in volume $\Delta V = V_{2} - V_{1} = 0.01 - 0.001 = 0.009 \ m^{3}$.
Substituting the values:
$W = -1 \times 10^{5} \times 0.009 = -900 \ J = -9 \times 10^{2} \ J$.

(D) According to the first law of thermodynamics,$\Delta U = Q + W$.
Given that the system absorbs heat,$Q = +8 \ kJ$.
Since the system does work on the surroundings,$W = -2.2 \ kJ$.
Substituting these values into the equation: $\Delta U = 8 \ kJ + (-2.2 \ kJ) = 5.8 \ kJ$.
Therefore,the internal energy change is $5.8 \ kJ$.

(B) For an isothermal reversible expansion of an ideal gas,the work done by the gas is given by:
$W = nRT \ln\left(\frac{V_2}{V_1}\right)$
Given:
$W = 5.187 \ kJ = 5187 \ J$
$V_1 = 10 \ m^3$
$V_2 = 20 \ m^3$
$T = 300 \ K$
$R = 8.314 \ J \ mol^{-1} \ K^{-1}$
Substituting the values:
$5187 = n \times 8.314 \times 300 \times \ln\left(\frac{20}{10}\right)$
$5187 = n \times 8.314 \times 300 \times 0.6931$
$5187 = n \times 1728.5$
$n = \frac{5187}{1728.5} \approx 3$
Therefore,the number of moles of gas used is $3$.

(B) According to the first law of thermodynamics,$\Delta U = Q + W$.
Given: Heat absorbed by the system,$Q = +4000 \ kJ$.
Work done on the system by the surroundings,$W = +2000 \ J = +2 \ kJ$.
Therefore,$\Delta U = 4000 \ kJ + 2 \ kJ = 4002 \ kJ$.

(A) The work done during the compression of a gas against a constant external pressure is given by the formula: $W = -P_{ext} \Delta V$.
Given:
$P_{ext} = 100 \ kPa = 10^{5} \ Pa$
$V_{initial} = 5 \ m^{3}$
$V_{final} = 2.5 \ m^{3}$
$\Delta V = V_{final} - V_{initial} = 2.5 \ m^{3} - 5 \ m^{3} = -2.5 \ m^{3}$
Substituting the values:
$W = -(10^{5} \ Pa) \times (-2.5 \ m^{3})$
$W = 2.5 \times 10^{5} \ J$
$W = 250,000 \ J = 250 \ kJ$.
Since the gas is compressed, work is done on the system, resulting in a positive value.

(D) Given: $n = 2 \text{ mol}$,$V_{1} = 5 \text{ m}^{3}$,$V_{2} = 10 \text{ m}^{3}$,$T = 300 \text{ K}$,$R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$.
For an isothermal reversible expansion,the work done is given by the formula: $W = -2.303 nRT \log_{10} \frac{V_{2}}{V_{1}}$.
Substituting the values: $W = -2.303 \times 2 \times 8.314 \times 300 \times \log_{10} \frac{10}{5}$.
$W = -2.303 \times 2 \times 8.314 \times 300 \times \log_{10} 2$.
Since $\log_{10} 2 \approx 0.3010$,we get: $W = -2.303 \times 2 \times 8.314 \times 300 \times 0.3010$.
$W \approx -3457.97 \text{ J} = -3.458 \text{ kJ}$.

(C) According to the first law of thermodynamics,$\Delta U = q + w$.
Since the process occurs at constant volume,the work done $w = -P \Delta V = 0$.
Given that heat is supplied to the system,$q = +500 \ J$.
Therefore,$q = 500 \ J$ and $w = 0$.

(C) Given: $V_{1} = 10 \ m^{3}$,$V_{2} = 20 \ m^{3}$,$T = 300 \ K$,$W = -5.187 \ kJ = -5187 \ J$.
For an isothermal reversible expansion,the work done is given by the formula:
$W = -2.303 \ nRT \log_{10} \left(\frac{V_{2}}{V_{1}}\right)$
Substituting the values:
$-5187 = -2.303 \times n \times 8.314 \times 300 \times \log_{10} \left(\frac{20}{10}\right)$
$-5187 = -2.303 \times n \times 8.314 \times 300 \times 0.3010$
$n = \frac{5187}{2.303 \times 8.314 \times 300 \times 0.3010}$
$n = \frac{5187}{1729.0} \approx 3 \ mol$.

(A) According to the first law of thermodynamics,$\Delta U = q + w$.
Here,work is done on the system,so $w = +2.5 \ kJ = +2500 \ J$.
The system releases heat,so $q = -1500 \ J$.
Therefore,$\Delta U = -1500 \ J + 2500 \ J = 1000 \ J$.

(C) The enthalpy of sublimation $(\Delta_{sub} H)$ is the sum of the enthalpy of fusion $(\Delta_{fus} H)$ and the enthalpy of vaporization $(\Delta_{vap} H)$.
Given:
$\Delta_{fus} H = 6.01 \ kJ \ mol^{-1}$
$\Delta_{vap} H = 45.07 \ kJ \ mol^{-1}$
For the process $H_2O_{(s)} \rightarrow H_2O_{(g)}$ at $0^{\circ}C$:
$\Delta_{sub} H = \Delta_{fus} H + \Delta_{vap} H$
$\Delta_{sub} H = 6.01 \ kJ \ mol^{-1} + 45.07 \ kJ \ mol^{-1} = 51.08 \ kJ \ mol^{-1}$

(A) The standard enthalpy of formation,$\Delta_{f} H^{\circ}$,is defined as the enthalpy change when $1 \text{ mole}$ of a compound is formed from its constituent elements in their standard states.
For $\Delta_{f} H^{\circ} = \Delta H^{\circ}$,the reaction must form exactly $1 \text{ mole}$ of the product from its elements in their standard states.
In option $A$,$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_{2}O_{(l)}$ represents the formation of $1 \text{ mole}$ of liquid water from its elements $H_2$ and $O_2$ in their standard states,thus $\Delta_{f} H^{\circ} = \Delta H^{\circ}$.

(D) Sublimation is the process of direct conversion of a solid into a gas.
According to Hess's Law,the enthalpy change for a process is the same whether it occurs in one step or multiple steps.
Solid to gas can be represented as:
$1. \text{Solid} \rightarrow \text{Liquid} \quad (\Delta_{fus} H)$
$2. \text{Liquid} \rightarrow \text{Gas} \quad (\Delta_{vap} H)$
Therefore,the total enthalpy change for sublimation is the sum of the enthalpy of fusion and the enthalpy of vaporization:
$\Delta_{sub} H = \Delta_{fus} H + \Delta_{vap} H$.

(C) Average bond enthalpy is the mean of the bond dissociation enthalpies of all bonds of the same type.
For water $(H_2O)$,the average bond enthalpy is given by $\Delta H_{avg} = \frac{\Delta H_1 + \Delta H_2}{2}$.
Given $\Delta H_{avg} = 464.5 \text{ kJ mol}^{-1}$ and $\Delta H_1 = 502 \text{ kJ mol}^{-1}$.
Substituting the values: $464.5 = \frac{502 + \Delta H_2}{2}$.
$929 = 502 + \Delta H_2$.
$\Delta H_2 = 929 - 502 = 427 \text{ kJ mol}^{-1}$.

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.
For the reaction: $2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)}$.
The change in the number of moles of gaseous species is calculated as: $\Delta n_g = \sum n_{p(g)} - \sum n_{r(g)}$.
$\Delta n_g = 2 - (2 + 1) = 2 - 3 = -1$.
Substituting this value into the equation,we get: $\Delta H = \Delta U + (-1) RT$.
Therefore,$\Delta H = \Delta U - RT$.

(A) The combustion reaction of benzene is:
$C_6H_{6(l)} + \frac{15}{2} O_{2(g)} \rightarrow 6 CO_{2(g)} + 3 H_2O_{(l)}$
The change in the number of gaseous moles is $\Delta n_g = n_{p(g)} - n_{r(g)} = 6 - 7.5 = -1.5$.
Given: $\Delta H = -3268 \ kJ \ mol^{-1}$,$T = 298 \ K$,$R = 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.
The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.
Rearranging for internal energy change (heat of combustion at constant volume): $\Delta U = \Delta H - \Delta n_g RT$.
Substituting the values: $\Delta U = -3268 - (-1.5 \times 8.314 \times 10^{-3} \times 298)$.
$\Delta U = -3268 + 3.716 = -3264.284 \ kJ \ mol^{-1}$.
Thus,the heat of combustion at constant volume is approximately $-3264.2 \ kJ \ mol^{-1}$.

(A) The molar mass of $O_2$ is $32 \ g/mol$.
Given mass of $O_2$ is $16 \ g$.
Number of moles of $O_2 = \frac{16 \ g}{32 \ g/mol} = 0.5 \ mol$.
Energy required to dissociate $0.5 \ mol$ of $O_2$ is $x \ kJ$.
Bond enthalpy is defined as the energy required to dissociate $1 \ mol$ of bonds.
Energy required for $1 \ mol$ of $O_2 = \frac{x \ kJ}{0.5 \ mol} = 2x \ kJ/mol$.
Therefore,the bond enthalpy of $O=O$ bond is $2x \ kJ$.

(B) The combustion reactions are:
$I. C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} \quad \Delta H_{1} = -x_{1}$
$II. H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_{2}O_{(l)} \quad \Delta H_{2} = -x_{2}$
$III. C_{2}H_{6(g)} + \frac{7}{2} O_{2(g)} \longrightarrow 2CO_{2(g)} + 3H_{2}O_{(l)} \quad \Delta H_{3} = -x_{3}$
We need the heat of formation of $C_{2}H_{6(g)}$:
$2C_{(s)} + 3H_{2(g)} \longrightarrow C_{2}H_{6(g)} \quad \Delta H_f = ?$
Perform the operation: $(2 \times I) + (3 \times II) - III$
$\Delta H_f = 2(-x_{1}) + 3(-x_{2}) - (-x_{3})$
$\Delta H_f = -2x_{1} - 3x_{2} + x_{3}$

(B) The given reaction is: $2 SO_{3(g)} \longrightarrow 2 SO_{2(g)} + O_{2(g)}$
Calculate the change in the number of moles of gaseous species,$\Delta n_g = \sum n_{g(products)} - \sum n_{g(reactants)}$
$\Delta n_g = (2 + 1) - 2 = 1$
Using the thermodynamic relation: $\Delta H = \Delta U + \Delta n_g RT$
Substituting $\Delta n_g = 1$,we get: $\Delta H = \Delta U + RT$
Therefore,$\Delta H - \Delta U = RT$

(C) The balanced chemical equation is $C(s) + 2S(s) \rightarrow CS_2(l) \quad \Delta H = 92 \ kJ \ mol^{-1}$.
This reaction is endothermic,meaning $92 \ kJ$ of heat is absorbed for every $1 \ mol$ $(12 \ g)$ of carbon reacted.
To find the heat for $6 \ g$ of carbon:
Heat $= (92 \ kJ \ mol^{-1} / 12 \ g \ mol^{-1}) \times 6 \ g = 46 \ kJ$.

(C) The target reaction is the hydration of ethene: $C_2H_{4_{(g)}} + H_2O_{(l)} \longrightarrow C_2H_5OH_{(l)}$.
We can obtain this by subtracting equation $(i)$ from equation $(ii)$:
$(ii) - (i): (C_2H_{4_{(g)}} + 3O_{2_{(g)}}) - (C_2H_5OH_{(l)} + 3O_{2_{(g)}})$ $\longrightarrow (2CO_{2_{(g)}} + 2H_2O_{(l)}) - (2CO_{2_{(g)}} + 3H_2O_{(l)})$
This simplifies to: $C_2H_{4_{(g)}} - C_2H_5OH_{(l)} \longrightarrow -H_2O_{(l)}$,which rearranges to $C_2H_{4_{(g)}} + H_2O_{(l)} \longrightarrow C_2H_5OH_{(l)}$.
The enthalpy change is $\Delta H^{\circ} = \Delta H^{\circ}_{(ii)} - \Delta H^{\circ}_{(i)} = -1410 \ kJ - (-1368 \ kJ) = -42 \ kJ$.

(B) The balanced chemical equation is: $2ClF_{(g)} + O_{2(g)} \longrightarrow Cl_2O_{(g)} + OF_{2(g)}$.
Given that $6.0 \ g$ of $O_2$ corresponds to $n = \frac{6.0 \ g}{32.0 \ g/mol} = 0.1875 \ mol$ of $O_2$.
The heat absorbed for $0.1875 \ mol$ of $O_2$ is $\Delta H = +38.55 \ kJ$.
The standard enthalpy of reaction $(\Delta H^0)$ is defined for the reaction as written,which involves $1 \ mol$ of $O_2$.
Therefore,$\Delta H^0 = \frac{38.55 \ kJ}{0.1875 \ mol} = 205.6 \ kJ/mol$.

(D) The thermochemical equation for the formation of water is:
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_{2}O_{(\ell)} \quad \Delta H_{f}^{0} = -286 \ kJ \ mol^{-1}$
This means $1 \ mol$ of $H_{2}O$ $(18 \ g)$ releases $286 \ kJ$ of energy.
Given mass of water $= 1800 \ mg = 1.8 \ g$.
Number of moles of $H_{2}O = \frac{1.8 \ g}{18 \ g \ mol^{-1}} = 0.1 \ mol$.
Energy liberated $= 0.1 \ mol \times 286 \ kJ \ mol^{-1} = 28.6 \ kJ$.

(D) According to the first law of thermodynamics,$\Delta U = Q + W$.
Given that heat is released from the system,$Q = -2 \ kJ$.
Since work is done on the system,$W = +6 \ kJ$.
Therefore,$\Delta U = -2 \ kJ + 6 \ kJ = +4 \ kJ$.

(C) The enthalpy change of a reaction is given by the sum of bond enthalpies of reactants minus the sum of bond enthalpies of products:
$\Delta H^{\circ} = \sum \Delta H^{\circ}_{(\text{reactants})} - \sum \Delta H^{\circ}_{(\text{products})}$
For the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$:
$\Delta H^{\circ} = [\Delta H^{\circ}_{(N \equiv N)} + 3 \Delta H^{\circ}_{(H-H)}] - [6 \Delta H^{\circ}_{(N-H)}]$
Substituting the given values:
$-83 = \Delta H^{\circ}_{(N \equiv N)} + 3(435) - 6(389)$
$-83 = \Delta H^{\circ}_{(N \equiv N)} + 1305 - 2334$
$-83 = \Delta H^{\circ}_{(N \equiv N)} - 1029$
$\Delta H^{\circ}_{(N \equiv N)} = 1029 - 83 = 946 \ kJ \text{ mol}^{-1}$

(B) The chemical equation for the formation of methane is: $C_{(s)} + 2H_{2(g)} \longrightarrow CH_{4(g)}$
The molar mass of graphite $(C)$ is $12.0 \ g \ mol^{-1}$.
Given that $6.0 \ g$ of graphite releases $37.4 \ kJ$ of heat,the enthalpy change $(\Delta H)$ for this amount is $-37.4 \ kJ$.
To find the standard enthalpy of formation $(\Delta H_f^{\circ})$,we calculate the heat released for $1 \ mol$ $(12.0 \ g)$ of graphite:
$\Delta H_f^{\circ} = \frac{-37.4 \ kJ}{6.0 \ g} \times 12.0 \ g \ mol^{-1} = -74.8 \ kJ \ mol^{-1}$.

(D) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.
For the reaction $N_{2(g)} + 3 H_{2(g)} \rightarrow 2 NH_{3(g)}$,the change in the number of moles of gaseous species $(\Delta n_g)$ is calculated as: $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 2 - (1 + 3) = 2 - 4 = -2$.
Substituting this value into the equation: $\Delta H = \Delta U + (-2) RT = \Delta U - 2 RT$.
Therefore,the correct option is $D$.

(A) The chemical formula of acetaldehyde is $CH_3CHO$. The molar mass of $CH_3CHO = (2 \times 12) + (4 \times 1) + 16 = 44 \ g \ mol^{-1}$.
Given that the heat of combustion for $1 \ mol$ $(44 \ g)$ of $CH_3CHO$ is $-1172 \ kJ \ mol^{-1}$.
For $66 \ g$ of $CH_3CHO$,the number of moles $n = \frac{66 \ g}{44 \ g \ mol^{-1}} = 1.5 \ mol$.
The heat liberated = $n \times \Delta H = 1.5 \ mol \times 1172 \ kJ \ mol^{-1} = 1758 \ kJ$.
Therefore,the amount of heat liberated is $1758 \ kJ$.

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.
Given that $75 \%$ is completed in $60 \ min$,the remaining concentration $[A]_t$ is $100 - 75 = 25 \%$ of $[A]_0$.
$k = \frac{2.303}{60} \log_{10} \frac{100}{25} = \frac{2.303}{60} \log_{10} 4 = \frac{2.303 \times 0.6020}{60} \approx 0.0231 \ min^{-1}$.
The time required for $50 \%$ completion is the half-life $t_{1/2}$.
$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0231} = 30 \ min$.

(A) According to the Arrhenius equation,the rate constant $K$ is given by $K = A e^{-E_a / RT}$.
When the activation energy $E_a$ decreases at a constant temperature $T$,the term $E_a / RT$ decreases.
Consequently,the exponent $-E_a / RT$ increases (becomes less negative).
Since the exponential function $e^x$ is an increasing function,$e^{-E_a / RT}$ increases,leading to an increase in the rate constant $K$.
Therefore,the term $E_a / RT$ decreases.

(A) The alkaline hydrolysis of tert-butyl bromide with aqueous alkali such as $NaOH$ or $KOH$ follows an $SN^1$ mechanism.
This is a two-step process:
Step $1$: $(CH_3)_3C-Br \rightarrow (CH_3)_3C^+ + Br^-$ (Slow,rate-determining step)
Step $2$: $(CH_3)_3C^+ + OH^- \rightarrow (CH_3)_3C-OH$ (Fast step)
In an energy profile diagram for a multi-step reaction,the highest peak corresponds to the transition state of the rate-determining step.
Since the first step is the rate-determining step,the highest energy barrier (peak) in the energy profile diagram represents the transition state of the $1^{st}$ step.

(B) The correct answer is $40 \%$.
Formalin is an aqueous solution of formaldehyde gas.
It typically contains $40 \%$ formaldehyde by volume in water.

(C) Blurring of vision is a side effect caused by the use of tranquilizers.
Other side effects include irregular sleep patterns,disorientation,confusion,restlessness,inability to relax,respiratory distress or arrest,and cardiac arrest.

(B) The chemical structure of $Cimetidine$ contains a cyano group $(-CN)$ attached to the nitrogen atom in the side chain.
Other options like $Ranitidine$ contain a nitro group $(-NO_2)$ and $Famotidine$ contains a sulfamoyl group $(-SO_2NH_2)$.
Therefore,$Cimetidine$ is the correct answer.

(D) $Aluminum$ $Hydroxide$ and $Magnesium$ $Hydroxide$ are common examples of antacids. They are prescribed to neutralize excess stomach acid,thereby preventing ulcers,relieving heartburn,and treating acid indigestion and stomach upsets.

(C) Disinfectants are chemical substances that kill or stop the growth of microorganisms but are not safe to be applied to living tissues.
$Bithional$ (also known as $Bithionol$) and $Chloroxylenol$ are used as antiseptics.
$Sulphur \ dioxide$ $(SO_2)$ is widely used as a disinfectant,a refrigerant,a reducing agent,a bleach,and a food preservative,especially in dried fruits.

(A) Sulfapyridine ($INN$; also known as sulphapyridine) is a sulfanilamide antibacterial medication.
It is classified as an antibiotic.
At one time,it was commonly referred to as $M\&B \ 693$.
Sulfapyridine is no longer prescribed for the treatment of infections in humans,but it may be used to treat linear $IgA$ disease and has applications in veterinary medicine.

(B) The compound $CH_3Br$ (methyl bromide) is commonly used as a fumigant for strawberries. However,among the given options,$CH_2Cl_2$ (dichloromethane) is often used as a solvent and in various industrial applications,but it is not the standard fumigant. Given the context of halogenated compounds,$CH_3Br$ is the correct chemical,but since it is not listed,the question likely refers to a specific context where $CH_2Cl_2$ is mentioned in certain older textbooks or specific industrial applications. Based on standard chemistry curricula,$CH_2Cl_2$ is the intended answer in this specific multiple-choice context.

(C) $2, 4-D$ or $2, 4$-dichlorophenoxy acetic acid is used as a herbicide.
It causes uncontrolled growth and selectively kills most broadleaf weeds.
It leaves most grasses such as cereals,lawn turf,and grassland unaffected.

(B) Aspirin is a non-narcotic analgesic,which means it relieves pain without causing addiction or sedation.
It is also an antipyretic,meaning it reduces body temperature.
Additionally,it has anti-blood clotting action,which is why it is often used to prevent heart attacks.
Therefore,the statement that aspirin belongs to narcotic analgesics is false.

(D) Bithionol (also known as Bithional) is added to soaps to impart antiseptic properties.
It acts as an antibacterial,antifungal,and algaecidal agent.
This helps in reducing the growth of microorganisms on the skin.

(A) Phenelzine is an antidepressant drug that acts as a monoamine oxidase inhibitor $(MAOI)$.
It treats depression by inhibiting the enzyme monoamine oxidase,which breaks down neurotransmitters like serotonin,norepinephrine,and dopamine.
By preventing their breakdown,it increases the concentration of these neurotransmitters in the brain,thereby improving mood and feelings of well-being.

(A) $o$-Acetylsalicylic acid is the chemical name for $Aspirin$.
It is a widely used medication that acts as an analgesic,antipyretic,and anti-inflammatory agent.

(C) $Na_2CO_3$ (sodium carbonate) is strongly alkaline and is not used as an antacid because it can be corrosive to the stomach lining.
Antacids are mild bases that neutralize excess stomach acid.
$Al(OH)_3$,$Mg(OH)_2$,and $NaHCO_3$ are commonly used as antacids to provide relief from acidity.

(C) Ethinylestradiol,commonly known as $Novestrol$,is a synthetic estrogen derivative.
It is widely used in birth control pills in combination with progestins.
It is also used as a component of menopausal hormone therapy to treat menopausal symptoms.

(A) Salvarsan is an organoarsenic compound with the chemical formula $C_{12}H_{12}As_2N_2O_2 \cdot 2HCl$.
By observing the molecular structure,we can identify the elements present:
$C$ (Carbon),$H$ (Hydrogen),$As$ (Arsenic),$N$ (Nitrogen),$O$ (Oxygen),and $Cl$ (Chlorine).
Phosphorus $(P)$ is not present in the structure of salvarsan.
Therefore,the correct option is $A$.

(C) Butylated hydroxyanisole $(BHA)$ is a well-known antioxidant.
It consists of a mixture of two isomeric organic compounds,$2-tert-butyl-4-hydroxyanisole$ and $3-tert-butyl-4-hydroxyanisole$.

(D) The correct answer is $D$.
An ideal drug should be specific to the affected site to minimize side effects.
Therefore,being 'not localised to the affected site' is $\underline{NOT}$ a characteristic of an ideal drug.
Key characteristics of an ideal drug include:
$1$. It should be effective against the target pathogen or disease.
$2$. It should be non-toxic and harmless to the host.
$3$. It should not disturb normal physiological processes.
$4$. It should be localised to the affected site to ensure maximum efficacy with minimal systemic toxicity.

(B) Phenelzine does not contain oxygen.
Phenelzine is a hydrazine derivative with the molecular formula $C_8H_{12}N_2$. It consists of a phenyl group attached to an ethyl chain,which is linked to a hydrazine moiety $(-NH-NH_2)$.
Serotonin $(C_{10}H_{12}N_2O)$,Veronal $(C_8H_{12}N_2O_3)$,and Iproniazid $(C_9H_{13}N_3O)$ all contain oxygen atoms in their structures.

(D) $0.2 \% $ solution of $Phenol$ is used as an antiseptic,while its $1 \%$ solution is used as a disinfectant.

(D) The correct answer is $D$.
$Sulphur \ dioxide$ is not an antiseptic drug.
Antiseptics are chemical substances that are applied to living tissues such as wounds,cuts,ulcers,and diseased skin surfaces to either kill or prevent the growth of microorganisms.
$Hydrogen \ peroxide$,$Bithional$,and $Chloroxylenol$ are well-known examples of antiseptics.

(C) Meprobamate is used to treat anxiety disorders or for short-term relief of the symptoms of anxiety in adults and children $6$ years of age and older.
Meprobamate belongs to a class of medications called tranquilizers.
It works by slowing down the activity in the brain to allow for relaxation.

(B) Codeine is an opioid analgesic that acts on the central nervous system.
It binds to opioid receptors in the brain and spinal cord,which leads to the depression of central nervous system functions,including respiration and pain perception.

(B) Tranquilizers are a class of chemical compounds used for the treatment of stress,and mild or even severe mental diseases.
$Equanil$,$Valium$,and $Veronal$ are well-known examples of tranquilizers.
$Novestrol$ is an estrogen derivative used as an antifertility drug,not a tranquilizer.

(D) Sucralose is the artificial sweetener that contains chlorine.
It is a trichloro derivative of sucrose.
It is $600$ times sweeter than cane sugar.

(C) Food preservation involves techniques that inhibit the growth of microorganisms or prevent chemical changes.
$1$. Removal of heat (refrigeration/freezing) slows down microbial growth.
$2$. Irradiation kills bacteria and parasites.
$3$. Addition of heat (pasteurization/canning) destroys enzymes and microorganisms.
$4$. Addition of water (hydration) actually promotes microbial growth and spoilage,so it is $\underline{\text{NOT}}$ a method of food preservation.

(C) $BHA$ stands for Butylated Hydroxyanisole.
It is a common food preservative.
The molecular formula of $BHA$ is $C_{11}H_{16}O_{2}$.

(B) $As_2O_3$,$P_2O_5$,and $SO_2$ are covalent compounds that exist as molecular solids or gases at room temperature,making them volatile.
$ZnO$ is an ionic solid with a high melting point and is non-volatile.

(C) The effective atomic number $(EAN)$ is calculated using the formula: $EAN = Z - X + Y$,where $Z$ is the atomic number of the metal,$X$ is the oxidation state of the metal,and $Y$ is the number of electrons donated by the ligands.
For $[Fe(CN)_6]^{4-}$:
$Z = 26$ (for $Fe$).
Let the oxidation state of $Fe$ be $x$. Then $x + 6(-1) = -4$,which gives $x = +2$. So,$X = 2$.
Since $CN^-$ is a monodentate ligand,$6$ ligands donate $6 \times 2 = 12$ electrons. So,$Y = 12$.
$EAN = 26 - 2 + 12 = 36$.

(B) The coordination number is defined as the total number of sigma bonds formed by the central metal atom with the ligands.
For monodentate ligands,the coordination number equals the number of ligands.
For polydentate ligands,the coordination number is the number of ligands multiplied by their denticity.
In $[Co(en)_3]^{3+}$,the ligand $en$ (ethylenediamine) is a bidentate ligand.
Therefore,the number of ligands is $3$,but the coordination number is $3 \times 2 = 6$.
Thus,the coordination number is not equal to the number of ligands.

(B) complex is homoleptic if the metal atom is bonded to only one type of donor atom/ligand. $A$ complex is cationic if the coordination sphere carries a positive charge.
$1$. $\left[CoCl_{2}(en)_{2}\right] Cl$: Heteroleptic (contains $Cl^-$ and $en$) and cationic.
$2$. $\left[Fe(H_{2}O)_{6}\right] Cl_{3}$: Homoleptic (only $H_{2}O$ ligands) and cationic $(\left[Fe(H_{2}O)_{6}\right]^{3+})$.
$3$. $K\left[Ag(CN)_{2}\right]$: Homoleptic but anionic $(\left[Ag(CN)_{2}\right]^{-})$.
$4$. $\left[Fe(CO)_{5}\right]$: Homoleptic but neutral.
Therefore,the correct option is $\left[Fe(H_{2}O)_{6}\right] Cl_{3}$.

(B) The chemical formula for hexamminecobalt$(III)$ chloride is $[Co(NH_3)_6]Cl_3$.
In this complex,the central metal ion is $Co^{3+}$.
The coordination number of $Co$ is $6$ (since there are $6$ $NH_3$ ligands).
The oxidation state of $Co$ is $+3$.
Since all ligands are the same $(NH_3)$,it is a homoleptic complex.
It is a cationic complex because the complex ion is $[Co(NH_3)_6]^{3+}$.
Therefore,the statement that the oxidation state of cobalt is $+3$ is true.

(B) complex is heteroleptic if it contains more than one type of ligand attached to the central metal atom.
It is cationic if the complex ion carries a positive charge.
In $[Co(NH_{3})_{4}Cl_{2}]Cl$,the coordination sphere is $[Co(NH_{3})_{4}Cl_{2}]^{+}$,which contains two different ligands ($NH_{3}$ and $Cl^{-}$) and carries a positive charge.
Thus,it is both heteroleptic and cationic.

(B) The $EAN$ (Effective Atomic Number) rule states that the total number of electrons around the central metal ion should equal the atomic number of the next noble gas.
$EAN = Z - \text{oxidation state} + 2 \times (\text{coordination number})$.
For $A: [Zn(NH_3)_4]^{2+}: EAN = 30 - 2 + 8 = 36$ (Obeys rule).
For $B: [Cu(NH_3)_4]^{2+}: EAN = 29 - 2 + 8 = 35$ (Does $NOT$ obey rule).
For $C: [Pt(NH_3)_6]^{4+}: EAN = 78 - 4 + 12 = 86$ (Obeys rule).
For $D: [Fe(CN)_6]^{4-}: EAN = 26 - 2 + 12 = 36$ (Obeys rule).
Therefore,the complex that does not obey the $EAN$ rule is $[Cu(NH_3)_4]^{2+}$.

(C) The formula for $EAN$ is $EAN = Z - X + Y$,where $Z$ is the atomic number,$X$ is the oxidation state,and $Y$ is the total number of electrons donated by ligands.
In $[Zn(NH_3)_4]SO_4$,the oxidation state of $Zn$ is $+2$ (since $NH_3$ is neutral and $SO_4$ is $-2$).
$Z = 30$,$X = 2$,and $Y = 4 \times 2 = 8$.
$EAN = 30 - 2 + 8 = 36$.

(B) tetradentate ligand is a ligand that can form $4$ coordinate bonds with a central metal atom or ion.
$1$. Ethylene diamine tetracetato $(EDTA^{4-})$ is a hexadentate ligand.
$2$. Triethylene tetramine $(trien)$ has $4$ nitrogen donor atoms,making it a tetradentate ligand.
$3$. Dimethyl glyoximato $(dmg^-)$ is a bidentate ligand.
$4$. Oxalato $(C_2O_4^{2-})$ is a bidentate ligand.
Therefore,the correct option is $B$.

(D) Dimethyl glyoximato (dmg) is a bidentate ligand with the structure $(CH_3C=NOH)_2$.
In the structure of dimethyl glyoxime,there are two $=N-OH$ groups attached to the carbon atoms.
Therefore,the number of $=N-OH$ groups present in dimethyl glyoximato is $2$.

(C) $EDTA$ stands for Ethylenediaminetetraacetate ion.
It contains two nitrogen atoms and four oxygen atoms as donor sites.
Therefore,it acts as a hexadentate ligand,meaning it has $6$ donor groups.

(A) An ambidentate ligand is a ligand that can coordinate to a central metal atom through two different donor atoms.
$NO_{2}^{-}$ is an ambidentate ligand because it can coordinate through either the nitrogen atom $(N)$ or the oxygen atom $(O)$.
$H_{2}O$ and $NH_{3}$ are monodentate ligands.
$C_{2}O_{4}^{2-}$ (oxalate ion) is a bidentate ligand.

(D) In the coordination compound $[CoCl_{2}(NH_{3})_{4}]Cl$,the central metal ion is $Co^{3+}$.
The ligands present inside the coordination sphere are $4$ $NH_{3}$ molecules and $2$ $Cl^{-}$ ions.
Therefore,the total number of ligands (coordination number) is $4 + 2 = 6$.

(B) monodentate ligand is a ligand that binds to a central metal atom through only one donor atom.
Water $(H_2O)$ is a monodentate ligand because it uses only the oxygen atom to donate a lone pair.
Oxalato $(C_2O_4^{2-})$,ethylenediamine $(en)$,and dimethylglyoximato $(dmg^-)$ are all bidentate ligands.

(D) The formula for $EAN$ is $EAN = Z - X + Y$,where $Z$ is the atomic number,$X$ is the oxidation state,and $Y$ is the number of electrons donated by ligands.
For $[Co(NH_3)_6]Cl_3$:
$1$. The oxidation state of $Co$ $(X)$ is $+3$.
$2$. The atomic number of $Co$ $(Z)$ is $27$.
$3$. The number of electrons donated by $6$ $NH_3$ ligands $(Y)$ is $6 \times 2 = 12$.
$EAN = 27 - 3 + 12 = 36$.

(D) The chemical formula of diethylene triamine (dien) is $NH_2-CH_2-CH_2-NH-CH_2-CH_2-NH_2$.
It contains $3$ nitrogen atoms,each having a lone pair of electrons that can be donated to a central metal ion.
Therefore,it acts as a tridentate ligand with $3$ donor groups.

(A) The coordination entity is Pentammine aqua cobalt$(III)$ iodide.
$NH_{3}$ (ammine) and $H_{2}O$ (aqua) are neutral ligands.
The oxidation state of $Co$ is $+3$.
To balance the charge of $+3$ on the complex cation $[Co(NH_{3})_{5}(H_{2}O)]^{3+}$,three iodide ions $(I^-)$ are required.
Therefore,the correct formula is $[Co(NH_{3})_{5}(H_{2}O)]I_{3}$.

(D) The oxalate ion $(C_{2}O_{4}^{2-})$ is a bidentate ligand with a charge of $-2$.
Given the oxidation state of $Al$ is $+3$.
Let the number of potassium ions be $x$.
For the complex to be neutral: $x(+1) + 1(+3) + 3(-2) = 0$.
$x + 3 - 6 = 0 \implies x = 3$.
Therefore,the formula is $K_{3}[Al(C_{2}O_{4})_{3}]$.

(A) $1$. Identify the ligands: There are two chloride ions $(Cl^-)$ named as 'dichlorido' and two ethylenediamine $(en)$ molecules named as 'bis(ethane-$1,2$-diamine)'.
$2$. Determine the oxidation state of the central metal atom $(Co)$: Let the oxidation state be $x$. The charge on $Cl$ is $-1$ and $en$ is neutral $(0)$. The overall charge on the complex is $+1$. So,$x + 2(-1) + 2(0) = +1$,which gives $x = +3$.
$3$. Assemble the name: Ligands are named alphabetically. 'Dichlorido' comes before 'bis(ethane-$1,2$-diamine)'.
$4$. The final name is Dichloridobis(ethane-$1,2$-diamine)cobalt$(III)$ ion.

(D) The name of the coordination compound is Barium tetrachlorocuprate$(II)$.
$1$. The cation is Barium,which is $Ba^{2+}$.
$2$. The anion is the coordination entity tetrachlorocuprate$(II)$.
$3$. 'Tetra' indicates four chloride ligands $(Cl^-)$,and 'cuprate$(II)$' indicates copper in the $+2$ oxidation state.
$4$. The charge on the complex ion $[CuCl_4]$ is calculated as: $x + 4(-1) = -2$,so $x = +2$. Thus,the complex ion is $[CuCl_4]^{2-}$.
$5$. Combining the cation $Ba^{2+}$ and the anion $[CuCl_4]^{2-}$,the formula is $Ba[CuCl_4]$.

(B) Let us analyze the charge on the complex ion for each option:
$A$: $[Co(en)_{3}]Cl_{3} \rightarrow [Co(en)_{3}]^{3+} + 3Cl^{-}$. The complex ion has a $+3$ charge.
$B$: $K_{3}[Al(C_{2}O_{4})_{3}] \rightarrow 3K^{+} + [Al(C_{2}O_{4})_{3}]^{3-}$. The complex ion has a $-3$ charge.
$C$: $[Ni(CO)_{4}]$ is a neutral complex with $0$ charge.
$D$: $[Ag(NH_{3})_{2}]Cl \rightarrow [Ag(NH_{3})_{2}]^{+} + Cl^{-}$. The complex ion has a $+1$ charge.
Therefore,the compound with a net negative charge on the complex ion is Potassium trioxalatoaluminate$(III)$.