Chemistry Test - 26

​​Homolysis and Heterolysis:

A complex has the molecular formula \(\mathrm{Co} .5 \mathrm{NH}_{3} . \mathrm{NO}_{2} . \mathrm{Cl}_{2}\). One mole of this complex produces three moles of ions in an aqueous solution. On reacting this solution with excess of \(\mathrm{AgNO}_{3}\) solution, we get 2 moles of white ppt. The complex is\(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right] \mathrm{Cl}_{2}\).

\(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right] \mathrm{Cl}_{2}\) can form three ions, one \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}\right]^{2+}\) and two \(\mathrm{Cl}^{-}\)ions.

As 1 mole of complex gives 2 moles of \(\mathrm{Cl}^{-}\)ions, 2 moles of AgCI will be precipitated.

Methyl salicylatehas a smell of oil of wintergreen.

• Methyl salicylate (oil of wintergreen or wintergreen oil) is an organic compound with the formula \(\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{OH})\left(\mathrm{CO}_{2} \mathrm{CH}_{3}\right)\).
• It is produced by many species of plants, particularly wintergreens.
• It is also produced synthetically.
• It is used as a fragrance, in foods and beverages, and in liniments.

\(\mathrm{BCl}_3\) and \(\mathrm{AlCl}_3\), both have vacant \(\mathrm{p}\)-orbital and incomplete octet, thus they behave as Lewis acids. \(\mathrm{SiCl}_4\) can accept lone pair of electron in d -orbital of silicon, hence it can act as Lewis acid. Although the most suitable answer is (d). However, both options (b) and (d) can be considered as correct answers. e.g. hydrolysis of \(\mathrm{SiCl}_4\)

i.e., option (b) \(\mathrm{AlCl}_3\) and \(\mathrm{SiCl}_4\) is also correct.

The electron-withdrawing inductive effect of the -\(OH\) group decreases the electron density on nitrogen, thus lowering the basicity of amines. This effect diminishes with distance from the amino group.

Thus, ethylamine \(> 3\) - amino - \(1\) - propanol \(> 2\) - amino ethanol

Germanium was discovered later which fit into the empty spaces left by Mendeleev and matched to the expected properties of "Eka-silicon". Chlorine, oxygen and silicon were included in Mendeleev's periodic table.

Germanium is a chemical element with the symbol Ge and atomic number 32. It is a lustrous, hard-brittle, grayish-white metalloid in the carbon group, chemically similar to its group neighbors silicon and tin. Pure germanium is a semiconductor with an appearance similar to elemental silicon.

Structural formula of Haloalkane isR-X.

It was derived fromalkanescontaining a halogens or more.

They are used as flame retardants, fire extinguishers, refrigerants, propellants, solvents, and medicinal items.

"RX"in which R is an alkyl or alkyl substituted group and X is a halogen (F, Cl, Br, I).

Haloalkanesare commonly used in organic synthesis as synthonous alternatives to alkyl cation (R+).

From ideal gas equation we know,

\(\mathrm{PV}=\mathrm{nRT} \)

\(\Rightarrow \mathrm{PV}=\frac{\mathrm{W}}{\mathrm{M}} \mathrm{RT} \)

\(\Rightarrow \mathrm{P}=\frac{\mathrm{W}}{\mathrm{V}} \cdot \frac{\mathrm{RT}}{\mathrm{M}} \)

\(\Rightarrow \mathrm{P}=\mathrm{d} \cdot \frac{\mathrm{RT}}{\mathrm{M}}\left[\because \mathrm{d}=\frac{\mathrm{W}}{\mathrm{V}}\right]\)

For a fixed amount of gas at a fixed temperature, \(\mathrm{M}\) and \(\mathrm{T}\) is constant.

\(\therefore \mathrm{P} \propto \mathrm{d}\)

So graph between pressure ( \(\mathrm{P}\) ) and density (d) is a straight line.

Also, \(\frac{\mathrm{P}}{\mathrm{d}}=\frac{\mathrm{RT}}{\mathrm{M}}\)

\(\Rightarrow \frac{\mathrm{P}}{\mathrm{d}} \propto \mathrm{T} \quad[\text { as } \mathrm{R}, \mathrm{M}=\text { constant }]\)

\(\therefore\) When temperature \(\mathrm{T}\) increases then slope of graph between \(\mathrm{P}\) and \(\mathrm{d}\) increases.

As \(\mathrm{T}_3>\mathrm{T}_2>\mathrm{T}_1\) then slope of \(\mathrm{T}_3\) will be highest then \(\mathrm{T}_2\) and then \(\mathrm{T}_1\).

So, graph (B) is the right graph.

We know the cell reaction

\(E_{\text {cell }}^{0}=E_{\text {Cathode }}^{0}-E_{\text {anode }}^{0}\)

Here Fe is cathode and \(\mathrm{A}\) is anode

\(\therefore E_{\text {cell }}^{0}=-0.44-(-1.66)\)

\(=-0.44+1.66\)

Cell reaction \(=+1.22 \mathrm{~V}\)

\(2 \mathrm{~A}-6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Al}^{3+}\) At anode

\(3 \mathrm{Fe}^{2+}+6 \mathrm{e}^{-} \rightarrow 3 \mathrm{Fe}\) At cathode

\(2 \mathrm{Al}+3 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Al}^{3+}+3 \mathrm{Fe}\) Cell reaction

Here

\(\mathrm{n}=6\) (Number of electron in cell reaction)

\(E_{\text {cell }}=E_{\text {cell }}^{0}+\frac{0.059}{n} \log \frac{\left[F e^{2+}\right]^{3}}{\left.[A]^{3+}\right]^{2}}\)

\(=1.22+\frac{0.059}{6} \log \frac{[0.02]^{3}}{[0.01]^{2}}\)

\(=1.22+9.8 \times 10^{-3} \log \left[\frac{8 \times 10^{-6}}{1 \times 10^{-4}}\right]\)

\(=1.22+9.8 \times 10^{-3} \times 1.25\)

\(=1.22+0.0122\)

\(=1.232 \mathrm{~V}\)

Weight of empty LPG cylinder \(=14.8 \mathrm{~kg}\)

Weight of full LPG cylinder \(=29 \mathrm{~kg}\)

\(\therefore\) Weight of gas \(=29-14.8=14.2 \mathrm{~kg}\)

If weight of full LPG cylinder \(=23 \mathrm{~kg}\)

then weight of gas used \(=29-23=6 \mathrm{~kg}\) at ambient temperature.

From ideal gas equaiton, \(\mathrm{pV}=\mathrm{nRT}\)

or \(\mathrm{pV}=\frac{\text { Weight of solute }}{\text { Molecular mass of solute }} \times \mathrm{RT}\)

\(\operatorname{orpV}=\frac{\mathrm{W}}{\mathrm{M}} \times \mathrm{RT}\)

Applying ideal gas to LPG cylinder when gas is full,

\(\mathrm{pV}=\mathrm{nRT} \)

\(3.47 \mathrm{~atm} \times \mathrm{V}=\frac{14.2 \mathrm{~kg}}{\mathrm{M}} \times \mathrm{RT} ...............\text{(i)}\)

Applying ideal gas to LPG cylinder when gas is reduced to \(23 \mathrm{~kg}\) at ambient temperature,

\(\mathrm{pV}=\mathrm{nRT} \)

\(\mathrm{p} \times \mathrm{V}=\frac{8.2 \mathrm{~kg}}{\mathrm{M}} \times \mathrm{RT}...............\text{(ii)}\)

Divide Eq. (i) by (ii)

\(\frac{3.47 \times \mathrm{V}}{\mathrm{p} \times \mathrm{V}}=\frac{\frac{14.2 \mathrm{~kg} \mathrm{RT}}{\mathrm{M}}}{\frac{8.2 \mathrm{~kg} \times \mathrm{RT}}{\mathrm{M}}} \)

\(\frac{3.47}{\mathrm{p}}=\frac{14.2}{8.2} \)

\(\Rightarrow \mathrm{p}=\frac{3.47 \times 41}{71}=2.003 \mathrm{~atm}\)

Hence, answer is 2.