NEET Chemistry Solved Question Paper 2009

At constant pressure,

  $$\begin{gathered} ∆\mathrm{p} =0 \\ ∆\mathrm{H}=∆\mathrm{E} +\mathrm{p}∆\mathrm{V} \mathrm{is} \mathrm{valid}. \end{gathered}$$

Second law of thermodynamics states that in a cyclic process heat cannot be converted completely into work because attainment of 0 K temperature is impossible.

Paschen series is the third series of hydrogen spectrum and lies in the infrared region, therefore for this series, n₁ = 3 and n₂ > n₁,ie, 4, 5, 6,....

$$\begin{gathered} \mathrm{Gibbs} \mathrm{free} \mathrm{energy} (∆{\mathrm{G}}^{°}) \mathrm{is} \mathrm{related} \mathrm{with} \mathrm{equilibrium} \mathrm{constant} \left(\mathrm{K}\right) \mathrm{as} \\ ∆{\mathrm{G}}^{°} = -\mathrm{RT} \ln \mathrm{K} \\ - \frac{∆{\mathrm{G}}^{°}}{\mathrm{RT}} = \ln \mathrm{K} \\ \mathrm{K} = {\mathrm{e}}^{\frac{-∆{\mathrm{G}}^{°}}{\mathrm{RT}}} \end{gathered}$$

Millimoles of 50 mL of 0.01 M HCl
             = 50 x 0.01 M = 0.5
pH of the solution =-log (0.5) = 0.3010

(a) On adding 5 mL of 1 M HCl,
           Concentration of the resultant solution

         =$\frac{50 \times 0.01+ 5 \times 1}{50+5}=0.1$

     :. pH of the solution = -log (0.1) = 1

(b) On adding 50 mL or 0.01 M HCl,
       Concentration of the resultant solution

         =$\frac{50 \mathrm{x} 0.01 + 50 \mathrm{x} 0.01}{50 + 50}= 0.01$

  :. pH of the solution = -log (0.01) = 2

(c) On adding 50 mL of 0.002 M HCl,
       concentration of the resultant solution

       =$\frac{50 \times 0.01+ 50 \mathrm{x} 0.002}{50 +50}= 6\times {10}^{-3}$

:. pH of the solution = -log (6 x 10^-3)= 2.22

(d) On adding Mg,

    $$\begin{gathered} \mathrm{Mg} +2\mathrm{HCl} \to {\mathrm{MgCl}}_2 + 2{\mathrm{H}}^{+} \\ \mathrm{Since}, \mathrm{pH} \infty \frac{1}{\left[{\mathrm{H}}^{+}\right]} \\ :. \mathrm{In} \mathrm{this} \mathrm{case} \mathrm{pH} \mathrm{decreases}. \end{gathered}$$

Frequency of one photon = 2500 s^-1
:. Frequency of 1 mole of photon
               = 6.023 x 10²³ x 2500 s^-1

$$\begin{gathered} \mathrm{Total} \mathrm{energy} \mathrm{of} 1 \mathrm{mol} \mathrm{of} \mathrm{photon}, \mathrm{E}=\mathrm{h\nu } \\ = 6.626 \times {10}^{-34}\mathrm{Js} \times 6.023 \times {10}^{23}\times 2500 {\mathrm{s}}^{-1} \\ = 9.97 \times {10}^{-7}\mathrm{J}\simeq 10 \mathrm{erg} \end{gathered}$$

Atomic mass of nitrogen= 14
Since the compound contains C, H and N, at
least one N atom must be there in the molecule.

          $$\begin{gathered} 20\% \mathrm{N} =14 \\ 100 \% =\frac{14}{20 }\times 100 = 70 \end{gathered}$$

:. The minimum molecular weight of the organic  compound = 70.

Heat of neutralisation= Heat of formation of water (1 mol).

$$\begin{gathered} \mathrm{HCl} + \mathrm{NaCl} \to \mathrm{NaCl} + {\mathrm{H}}_2\mathrm{O}; ∆\mathrm{H} =-\mathrm{x} \\ \mathrm{or}\underset{1\mathrm{mol} }{ {\mathrm{H}}^{+}}+\underset{1\mathrm{mol}}{ {\mathrm{OH}}^{-} }\to {\mathrm{H}}_2\mathrm{O}; ∆\mathrm{H} = \mathrm{x} \\ {\mathrm{H}}_2{\mathrm{SO}}_4 + 2\mathrm{NaOH} \to {\mathrm{Na}}_2{\mathrm{SO}}_4 + 2{\mathrm{H}}_2\mathrm{O} ; ∆\mathrm{H} = -\mathrm{y} \\ \mathrm{or} 2{\mathrm{H}}^{+}+ 2{\mathrm{OH}}^{-}\to 2{\mathrm{H}}_2\mathrm{O}; ∆\mathrm{H} =-\mathrm{y} \\ {\mathrm{H}}^{+} + {\mathrm{OH}}^{-} \to {\mathrm{H}}_2\mathrm{O}; ∆\mathrm{H} =-\frac{\mathrm{y}}{2} \\ ∆\mathrm{H} = ∆\mathrm{H} \\ \mathrm{x} =\frac{\mathrm{y}}{2} \end{gathered}$$

Number of hybrid orbitals = number of $\mathrm{\sigma }$ bonds + number of lone pairs

       $\Rightarrow 3\mathrm{\sigma }$ bonds + 1 lone pair = 4

.:. Hybridisation of N is sp³ and geometry of the molecule is pyramidal.

$$\begin{gathered} \mathrm{In} \mathrm{the} \mathrm{relation}, {\mathrm{K}}_{\mathrm{p}}= {\mathrm{K}}_{\mathrm{c}}(\mathrm{RT}{)}^{\mathrm{x}}, \\ \mathrm{x}= {\mathrm{n}}_{\mathrm{p}}-{\mathrm{n}}_{\mathrm{r}} \\ \mathrm{For} \mathrm{the} \mathrm{reaction}, \\ {\mathrm{SO}}_2(\mathrm{g}) + \frac{1}{2}{\mathrm{O}}_2(\mathrm{g}) \rightleftharpoons {\mathrm{SO}}_3(\mathrm{g}) \\ \mathrm{x} = 1 - \left(1 +\frac{1}{2}\right) =-\frac{1}{2} \end{gathered}$$