Chemistry Test - 37

Techniques like titration, precipitation, spectroscopy, chromatography, etc. commonly used in analytical chemistry.

A branch of chemistry that deals with the identification of compounds and mixtures (qualitative analysis) or the determination of the proportions of the constituents (quantitative analysis): techniques commonly used are titration, precipitation, spectroscopy, chromatography, etc.

Adsorption takes place only when \(\Delta\)G \(<0\) i.e., \(\Delta\)H - T \(\Delta\)S \(<0\).

Where, \(\Delta\)G is Gibb's energy, \(\Delta\)H is enthalpy, \(\Delta\)S is entropy and T is Temperature.

So, if \(\Delta\)H is positive and \(\Delta\)S is negative then \(\Delta\)H-T \(\Delta\)S will always be positive. So, adsorption will not take place.

When pressure on piece of ice is increases, its melting point decreases.

The melting point of ice decreases when pressure increases because when pressure is increased volume is decreased and volume of water is less than ice.

For most substances, increasing the pressure when a system is in equilibrium between liquid and solid phases will increase the phase transition temperature. Water is one of a few special substances for which the pressure lowers the temperature of transition. The basic reason is that water actually expands when it goes from the liquid to solid phase.

In diamond, each C -atom is covalently bonded to four other Catoms to give a tetrahedral unit, so it shows \(\mathrm{sp}^3\) hybridisation. Therefore, each \(\mathrm{C}\)-atom forms four sigma bonds with neighbouring \(\mathrm{C}\)-atoms. In diamond each \(\mathrm{C}\)-atom utilizes its four unpaired electrons in bond formation. These bonding electrons are localized. Due to this reason diamond is a bad conductor of electricity. In graphite each C-atom is covalently bonded to three Catoms to give trigonal geometry. Each \(\mathrm{C}\)-atom in graphite is \(\mathrm{sp}^2\) hybridized. Three out of four valence electrons of each \(\mathrm{C}\)-atom are used in bond formation while the fourth electron is free to move in the structure of graphite. Due to this reason graphite is a good conductor of electricity.

For the reaction,

\(\underset{1}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3(\mathrm{~g})}\underset{5}{+5 \mathrm{O}_2(\mathrm{~g})} \rightarrow \underset{3}{3 \mathrm{CO}_2(\mathrm{~g})}\underset{4}{+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g})}\)

1 mole of propane reacts completely with 5 moles of oxygen to form 3 moles of carbon dioxide and 4 moles of steam.

\(44 \mathrm{~g}\) of propane \(=1\) mole of propane

\(100 \mathrm{~g}\) of propane \(=\frac{1}{44} \times 100=2.27 \mathrm{~mol}\)

\(32 \mathrm{~g}\) of \(\mathrm{O}_2=1\) mole of \(\mathrm{O}_2\)

\(1000 \mathrm{~g}_{\mathrm{g}}\) of \(\mathrm{O}_2=\frac{1}{32} \times 1000=31.25\) moles

\(\therefore 2.27\) moles of propane requires \(5 \times 2.27=11.35\) moles of \(\mathrm{O}_2\)

moles of \(\mathrm{CO}_2\) formed \(=3 \times 2.7=6.681 \mathrm{~mol}\) of \(\mathrm{CO}_2\)

When reaction is completed 19.90 moles of \(\mathrm{O}_2\),

6.81 moles of \(\mathrm{CO}_2\) and 9.08 moles of steam are left in the flask.

\(\text { Mole fraction of } \mathrm{CO}_2=\frac{\text { Moles of } \mathrm{CO}_2(\mathrm{~g})}{\text { Total number of moles }} \)

\(=\frac{6.81}{19.90+6.81+9.08}=0.19 \)

\(x \times 10^{-2}=0.19 \)

\( x=19\)

One \(\mathrm{Fe}^{3+}\) ion replaces three \(\mathrm{Na}^{+}\)ions. As \(\mathrm{Na}^{+}\)ions occupy all octahedral voids of \(\mathrm{NaCl}\) crystal, due to replacement few octahedral voids fall vacant. So, one \(\mathrm{Fe}^{3+}\) ion creates two vacant octahedral voids.

\(10^{-3}\) mole \(\mathrm{FeCl}_{3}\) creates unoccupied octahedral voids \(=2 \times 6.02 \times 10^{23} \times 10^{-3}=12.04 \times 10^{20}\)

So, \(100\) mole of doped \(\mathrm{NaCl}\) crystals create \(12.04 \times 10^{20}\) vacant octahedral voids. Hence, \(1\) mole of doped \(\mathrm{NaCl}\) crystals create \(12.04 \times 10^{18}\) vacant octahedral voids.

Depression in freezing point is a colligative property which depends upon the amount of the solute.

\(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \mathrm{m}\)

Thus, for a given solvent and given concentration, \(\Delta T_{f}\) is directly porportional to \(i\) (Van't Hoff factor) i.e. maximum \(T_{f}\) (and hence lowest freezing point) will correspond to maximum value of \(i \).

(A) \(A l_{2}\left(S O_{4}\right)_{3} \stackrel{H_{2} O}{\longrightarrow} 2 A l^{3+}+3 S O_{4}^{2-}\)

Here, Van't Hoff factor, i = 5

(B) \(C_{5} H_{10} O_{8} \stackrel{H_{2} O}{\longrightarrow}\) No ionization

Here, Van't Hoff factor, i = 1

(C) \(K I \stackrel{H_{2} O}{\longrightarrow}K^{+}+I^{-}\)

Here, Van't Hoff factor, i = 2

(D) \(C_{12} H_{22} O_{11} \stackrel{H_{2} O}{\longrightarrow}\) No ionization

Here, Van't Hoff factor, i = 2

Therefore, \(0.10 M(\approx 0.10 \mathrm{~m})\) aqueous solution will have the lowest feezing point.

Gallium remains liquid up to \(2276\) Kelvin.

Low melting point of gallium is due to the fact that it consists of \(Ga_{2}\) molecules and gallium remains liquid up to \(2276\) \(k\). Hence it is used in high-temperature thermometer. Gallium as a chemical symbol that is \(Ga\) and its atomic number is given as \(31\).

Bond order \(=\frac{\left(N_{ b }-N_{ a }\right)}{2}\), so the molecular orbital configuration of \(O _2^{-}\) is, \(\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^2=\pi^* 2 p_y^1\)

Bond order of \(O _2^{-}=(\frac{1}{2})(10-7)=1.5\).

The molecular orbital configuration of \(O _2^{2-}\) is \(\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^2=\pi^* 2 p_y^2\)

Bond order of \(O _2^{2-}=(\frac{1}{2})(10-8)=1\).

The molecular orbital configuration of \(O _2^{2+}\) is \(\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2\)

Bond order of \(O _2^{2+}=(\frac{1}{2})(10-4)=3\). 

The molecular orbital configuration of \(O _2^{+}\), \(\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi * 2 p_x^1\)

Bond order of \(O _2^{+}=(\frac{1}{2})(10-5)=2.5\). 

Out of these, \(O _2^{2+}\) has the highest bond order, so the lowest bond length as bond length is inversely proportional to bond order.