NEET Chemistry Solved Question Paper 2009

Equivalent weight of metal carbonate = 20 + 30 = 50

2.5 gm of metal carbonate = $\frac{2.5}{50}$= 0.05 eq.

Number of equivalent of H₂SO₄ would have reacted = 0.05

Number of equivalent of H₂SO₄ taken = $\frac{100 \times 1}{1000}$= 0.1

Number of equivalent weight of H₂SO₄ reamins unreacted = 0.1 - 0.05 = 0.05 eq.

$\therefore$ Number of equivalent of alkali consumed = 0.05 eq.

milli eq = Normality $\times$ Volume in mL

$\therefore$1.0 $\times$ V = 0.05 $\times$ 1000

V = $\frac{0.05 \times 1000}{1.0}$ = 50 mL

pH of a 0.01 M solution is 

$$\begin{gathered} \mathrm{\alpha } = \sqrt{\frac{{\mathrm{K}}_{\mathrm{a}}}{\mathrm{C}}} = \sqrt{\frac{6.6 \times {10}^{-4}}{0.01}} = \sqrt{6.6 \times {10}^{-2}} = 0.257 \\ \left[{\mathrm{H}}^{+ }\right] = \mathrm{C\alpha } = 0.01 \times 0.257 = 2.57 \times {10}^{-3} \end{gathered}$$

pH = 3 - log2.5 = 2.60

For n moles of an ideal gas PV = nRT

or V = n$\frac{RT}{P}$

Differentiating  with respect to T at constant P, we have

$$\begin{gathered} {\left(\frac{dV}{dT}\right)}_P = \frac{nR}{P} = \frac{V}{T} \\ \mathrm{\alpha } = \frac{1}{V}{\left(\frac{dV}{dT}\right)}_P = \frac{1}{V} \times \frac{V}{T} = \frac{1}{T} \end{gathered}$$

Therefore, for an ideal gas $\mathrm{\alpha }$ is equal to $\frac{1}{T}$.

Mercurous chloride or calomel mineral is a rare mineral. It is dense white or yellowisg- white, odorless solid. It is a component of reference electrodes in electrochemistry. It exists in the form of Hg₂²⁺.

Molarity = $\frac{\mathrm{weight} \mathrm{of} \mathrm{solute}}{\mathrm{molecular} \mathrm{weight}} \times \frac{1000}{\mathrm{volume} \mathrm{of} \mathrm{solution} (\mathrm{in} \mathrm{mL})}$

(Since, Volume of solution = $\frac{\mathrm{mass}}{\mathrm{density}}$)

Therefore, volume of solution = $\frac{100}{1.84}$ = 54.34 mL

Molarity = $\frac{98}{98} \times \frac{1000}{54.34} = 18.4 \mathrm{M}$

In OF₂ repulsion between lone pairs is greater than that between bond pair since electrons are away from O and nearer to F. In Cl₂O, bonding electrons are nearer to O than to Cl, so the bond angle is greater than 109°28'. In Br₂O, the bonding electrons are more closer to oxygen than in Cl₂O, so the bond angle is largest (116°).

The fluoride ion, by a process of hydrogen bonding, forms the anion, HF${}_2^{-}$. The compound is written as K⁺[HF₂]^-.

SeF₄ and XeO₂F₂ are both sp³ hybridized, trigonal bipyramidal and see-saw shaped with 1 lone pair of electrons each.

SF₄ has 1 lone pair, XeF₂ has 3 lone pairs.

XeOF₄ is square pyramidal with 1 lone pair. TeF₄ is see-saw shaped with 1 lone pair.

SeCl₄ has 1 lone pair and is see-saw shaped. XeF₄ has 2 lone pairs and is planar in shape.

$$\begin{gathered} \underset{\mathrm{At} \mathrm{equilibrium}}{\underset{\mathrm{Initial}}{}} \underset{1 - \mathrm{\alpha }}{\underset{1}{\mathrm{\alpha }-\mathrm{D} \mathrm{glucose}}} \rightleftharpoons \underset{\mathrm{\alpha }}{\underset{0}{\mathrm{\beta }-\mathrm{D} \mathrm{glucose}}} \\ \mathrm{K} = \frac{\mathrm{\alpha }}{1 - \mathrm{\alpha }} \end{gathered}$$

Solving $\mathrm{\alpha }$ = 0.642

1 - $\mathrm{\alpha }$ = 0.358

Percent of $\mathrm{\alpha }$-D glucose remaining at equilibrium = 35.8%

$\underset{\mathrm{Equilibrium}}{\underset{\mathrm{Initial} \mathrm{concentration}}{{}_{}}} \underset{0.4 \mathrm{mole}}{\underset{1.0 \mathrm{mole}}{2{\mathrm{SO}}_{3 \left(\mathrm{g}\right)}}} \rightleftharpoons \underset{0.6 \mathrm{mole}}{\underset{0.0 \mathrm{mole}}{2{\mathrm{SO}}_{2 \left(\mathrm{g}\right)}}} + \underset{0.3 \mathrm{mole}}{\underset{0.0 \mathrm{mole}}{{{\mathrm{O}}_2}_{ \left(\mathrm{g}\right)}}}$

Equilibrium constant is given by

K = $\frac{{\left[{\mathrm{SO}}_2\right]}^{2 }\left[{\mathrm{O}}_2\right]}{{\left[{\mathrm{SO}}_3\right]}^2} = \frac{{\displaystyle {\left(0.6\right)}^2}{\displaystyle }{\displaystyle \left(0.3\right)}}{{\displaystyle {\left(0.4\right)}^2}} = 0.675 \approx 0.68$