Chemistry Test-12

Compressibility factor \(Z\) is given by

\(Z=\frac{PV}{R T}\)

Where \(P\) is pressure, \(V\) is the molar volume of the gas, \(R\) is the Universal gas constant, \(T\) is the temperature.

At critical point:

\(P=\frac{a}{27 b^{2}}\)

\(V =3 b\)

\(T=\frac{8 a}{27 R b}\)

\(Z=\frac{P V}{R T}\)

Putting the value of \(P, V, T\) in \(Z\).

\(Z=\frac{\frac{a}{27 b^{2}} \times 3 b}{R \times \frac{8 a}{27 R b}}=\frac{3}{8}=0.375\)

Consider the following equation:

\(\left[\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{O}_{2}\right] \times 2\)

\(2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\quad \mathrm{O}_{2}\)

\(68 \mathrm{g}\) \(22.4 \mathrm{L}\) at \(\mathrm{NTP}\)

\(\because 22.4 \mathrm{L} \mathrm{O}_{2}\) at \(\mathrm{NTP}\) is obtained by \(68 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\)

\(\therefore 20 \mathrm{L} \mathrm{O}_{2}\) at \(\mathrm{NTP}\) will be obtained by \(\mathrm{H}_{2} \mathrm{O}_{2}\)

\(\therefore 1000 \mathrm{mL} \mathrm{O}_{2}\) at \(\mathrm{NTP}\) is obtained by \(\mathrm{H}_{2} \mathrm{O}_{2}=60.7 \mathrm{g}\)

Percentage strength \(=\frac{60.7 \times 100}{1000}=6.07\)

As valency of phosphorus is 5 and there are in total four phosphorus atoms. Each \(P\) atom has one lone pair of electron and each covalent bond (includes two electrons) will be formed Therefore, in total there will be 4 lone pair of electrons and six P-P single bonds in a molecule of white phosphorus.

\(P_4\) has six \(P− P\) single bonds.

Properties of Phosphorus \(\left(P_{4}-\text { White }\right)\)

White or slightly yellow, waxy, non-conductor of heat and electricity, insoluble in water but soluble in many organic solvents.

It is highly reactive owing to the large angle strain in the tetrahedral structure of \(P_{4}\).

Allotropy of elements of nitrogen family group (VA) Allotropy: All the members of group \(15\) except \(Bi\) exhibit the phenomenon of allotropy.

Nitrogen exists in two solid and one gaseous allotropic forms.

Phosphorus exists in several allotropic forms such as white, red, scarlet, violet and black form.

a. White or yellow phosphorus: White phosphorus is prepared from rock phosphate \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}, \mathrm{SiO}_{2}\) and coke which is electrically heated in a furnace.

\(2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}+6 \mathrm{SiO}_{2} \stackrel{\Delta}{\rightarrow} 6 \mathrm{CaSiO}_{3}+\mathrm{p}_{4} \mathrm{O}_{10}\)

\(\mathrm{P}_{4} \mathrm{O}_{10}+10 \mathrm{C} \stackrel{\Delta}{\rightarrow} \mathrm{P}_{4}+10 \mathrm{CO}\)

When exposed to light, it acquires a yellow colour.

b. Red phosphorus: It is obtained by heating yellow phosphorus, between \(240-250^{\circ} \mathrm{C}\) in the presence of inert gas. Yellow phosphorus can be separated from red phosphorus by reaction with \(\mathrm{NaOH}\) (aq) or \(\mathrm{KOH}\) (aq) when the former reacts and the latter remains unreacted.

Arsenic exists in three allotropic forms namely grey, yellow and black. Antimony also exists in three forms, viz, metallic, yellow and explosive.

The square planar complex of the type \(MAXBL\) (where \(A, B, X\), and \(L\) are unidentate ligands) shows two cis and one trans isomer as shown in the image.

The isomeric hexane should have two different types of H atoms and four similar types of H atoms to give two monochlorinated products.

In 2,3-dimethyl butane, H atoms at positions 'a' and 'b' are the same and 'b' are the same. So, two positions are available for chlorination.

Lactose is more commonly known as milk sugar since this disaccharide is found in milk.

Lactose is composed of \(\beta-D-g a l a c t o s e\) and \(\beta-D-g l u c o s e .\)The linkage is between C1 of galactose and C4 of glucose.Free aldehyde group may be produced at C-1 of glucose unit, hence it is also a reducing sugar.

Another disaccharide, maltose is composed of two a-D-glucose units in which C1 of one glucose (I) is linked to C4 of another glucose unit (II). The free aldehyde group can be produced at C 1 of second glucose in solution and it shows reducing properties so it is a reducing sugar.

Thermosetting polymers is the type of polymers is Bakelite.

• Bakelite monomer is phenol and formaldehyde.
• Due to its excellent insulating properties, it is used for making switches.
• High resistance to electricity and heat.

Given:

\(k=\frac{0.693}{{ }_{t_{\frac{1}{2}}}}\)

\(=\frac{0.693}{5730} \text { years }{ }^{-1}\)

It is known that,

\( t =\frac{2.303}{k} \log \frac{[R]_{0}}{[R]} \)

\(=\frac{2.303}{\frac{0.693}{5730}} \log \frac{100}{80} \)

\(=1845 \text { years } \text { (approximately) }\)

An emulsifier is a substance stabilises the emulsion by increasing its kinetic stability.

In egg yolk the main emulsifying agent is lecithin. Other examples of emulsifiers include soy lecithin, sodium phosphates and sodium stearoyl lactylate.

Phenol gets slowly oxidised to a pink coloured compound p-benzoquinone when exposed to air.

This is due to the formation of o-and p-benzoquinone

s-block of the periodic table the element with atomic number 56 belongs.

Barium (Ba):

• The atomic number is 56.
• It is mostly used in drilling fluids for oil and gas wells and also used in paints and glassmaking.
• It is found in combination with other elements.
• Barite is one of the major ores of Barium.
• Atomic number 56, Barium belongs to s - block elements.
• Modern Periodic Table is classified into the following elements - p-block, s-block, d-block, and f-block.
• s-block elements include alkali metals and alkaline earth metals.

The balanced chemical neutralization reaction of phosphoric acid with calcium hydroxide is as follows.

\(2 \mathrm{H}_{3} \mathrm{PO}_{4}+3 \mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}+6 \mathrm{H}_{2} \mathrm{O}\)

In the above chemical reaction 2 moles of phosphoric acid reacts with three moles of calcium hydroxide and forms one mole of calcium phosphate and six moles of water as the products.

Complete Solution :

In the question it is given that to find the volume of \(0.025 \mathrm{M}\) phosphoric acid to neutralize \(25 \mathrm{ml}\) of \(0.003 \mathrm{M}\) calcium hydroxide.

The neutralization reaction of phosphoric acid with calcium hydroxide is as follows.

\(2 \mathrm{H}_{3} \mathrm{PO}_{4}+3 \mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}+6 \mathrm{H}_{2} \mathrm{O}\)

• Number of moles of phosphoric acid in the above reaction = 2 moles.
• Number of moles of calcium hydroxide in the above reaction \(=3\) moles
• There is a formula to calculate the number of moles of \(0.025 \mathrm{M}\) phosphoric acid and it is as follows.
• Number of moles \(=(\) Volume in Liter \()\) (Molarity \()\)
• In the question it is given that the volume of calcium hydroxide is \(25 \mathrm{ml}=0.025 \mathrm{~L}\)
• Number of moles of calcium hydroxide \(=(0.025 \mathrm{~L})(0.03)=\) \(7.5 \times 10^{-4}\) moles
• Number of moles of phosphoric acid required to react with
• \(7.5 \times 10^{-4}\) moles of calcium hydroxide \(=\)
• \(\frac{2}{3} \times 7.5 \times 10^{-4} \text { moles }=5 \times 10^{-4}\)
  
• So \(5 \times 10^{-4}\) moles phosphoric required to react with calcium hydroxide.

Then the volume of the phosphoric acid is\(\frac{5 \times 10^{-4}}{0.025}=0.02 L=20 m l\)

The element Californium belongs to the family of Actinide series. In the periodic table, Actinoid element or the actinide element, any of a series of 15 consecutive chemical elements from actinium to lawrencium (atomic numbers 89–103).

• As a group, Actinoid elements are significant largely because of their property of being radioactive.
• Curium (Cm), a synthetic chemical element having atomic number 96 comes under the Actinoid series of the periodic table.
• Actinoid elements are actinium, thorium, protactinium, uranium, neptunium, plutonium, americium, curium, berkelium, californium, einsteinium, fermium, mendelevium, nobelium and lawrencium.