Chemistry Test-1

For \(\mathrm{NaCl}\)

\(\mathrm{pV}=\mathrm{nST} \times \mathrm{i}\)

\(\mathrm{p}_{\mathrm{1}} \times \frac{100}{1000}=\frac{1.2}{58.5} \times 0.0821 \times \mathrm{T} \times \mathrm{i} \ldots\) (i)

For glucose, \(\mathrm{pV}=\mathrm{nST}\)

\(\mathrm{p}_{2} \times \frac{100}{1000}=\frac{7.2}{180} \times 0.0821 \times \mathrm{T} \ldots\) (ii)

\(\because \mathrm{NaCl}\) and glucose solutions are isotonic

\(\therefore \mathrm{p}_{1}=\mathrm{p}_{2}\)

On dividing Eq. (i) by (ii), we have

\(\mathrm{i}=\frac{7.2}{180} \times \frac{58.5}{1.2}=1.95=2\)

Basicity increases in the presence of \(+1\) groups and decreases in the presence of \(-1\) groups.\(\mathrm{CN}\) is electron withdrawing group (-I group) and \(\mathrm{CH}_{3}, \mathrm{NH}_{2}\) are electron releasing groups \((+1)\).

Thus, the correct order of basicity is (IV)<(I)<(III)<(II).

According to the first law of thermodynamics,

\(\Delta \mathrm{U}=\mathrm{q}+\mathrm{W} \ldots .\) (i)

Where,

\(\Delta \mathrm{U}=\) change in internal energy for a process

\(\mathrm{q}=\) heat

\(\mathrm{W}=\) work

Given,

\(\mathrm{q}=+701 \mathrm{~J}\) (Since heat is absorbed)

\(\mathrm{W}=-394 \mathrm{~J}\) (Since work is done by the system)

Substituting the values in expression (i), we get

\(\Delta \mathrm{U}=701 \mathrm{~J}+(-394 \mathrm{~J})\)

\(\Delta \mathrm{U}=307 \mathrm{~J}\)

Hence, the change in internal energy for the given process is \(307 \mathrm{~J}\).

The \(\mathrm{C}-\mathrm{Cl}\) bond in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}\) is shorter than in \(\mathrm{CH}_{3} \mathrm{Cl}\) and therefore, stronger.

• The carbon in \(\mathrm{C-Cl}\) bond in chlorobenzene is \(\mathrm{sp}^{2}\) hybridised, while in \(\mathrm{CH} 3-\mathrm{Cl}\) is \(\mathrm{sp}^{3}\) hybridised.
• In \(\mathrm{sp}^{2}\) hybrid orbitals have more of \(\mathrm{s}\)-character and hence the carbon if chlorobenzene withdraws the electron pair between \(\mathrm{C-Cl}\) with greater force.
• As a result, \(\mathrm{C-Cl}\) bond is shorter than \(\mathrm{CH}_{3}-\mathrm{Cl}\).
  
• The lone pair on chlorine are dispersed throughout the benzene ring by resonance.
• This gives the \(\mathrm{C-Cl}\) bond a double bond character.

Trans effect can be defined as the effect of a ligand towards substitution on other ligand which is trans to it.

\(\mathrm{F}^{-}<\mathrm{HO}^{-}<\mathrm{H}_{2} \mathrm{O}<\mathrm{NH}_{3}<\mathrm{py}<\mathrm{Cl}^{-}<\mathrm{Br}^{-}<\mathrm{I}^{-}<\mathrm{SCN}^{-}<\mathrm{NO}_{2}^{-}<\mathrm{SC}\left(\mathrm{NH}_{3}\right)_{2}<\mathrm{Ph}^{-}<\mathrm{SO}_{3}^{2-}\)

\(<\mathrm{PR}_{3}<\mathrm{AsR}_{3}<\mathrm{SR}_{2}<\mathrm{H}_{3} \mathrm{C}^{-}<\mathrm{H}^{-}<\mathrm{NO}_{2}<\mathrm{CO}^{-}<\mathrm{NC}_{2} \mathrm{H}_{4}\)

Trans effect is nothing but those ligands which are trans directing. That means, which prefer a trans structure. Mostly in square planar complexes.

D-Glucose and D-Fructose are the hydrolysis products of sucrose.

Sucrose is a saccharose, also known by names such as table sugar, cane sugar, beet sugar, etc. It is a carbohydrate, more specifically, a disaccharide made up of two monosaccharides, namely, glucose and fructose.

Glucose is an aldehyde, which forms a six-carbon ring. Whereas, fructose is a ketone, which forms a five-membered ring (so, a furan). Therefore, on hydrolysis sucrose will break into its monomeric sugars. The reaction can be represented as:

Sucrose is made by linking C1 of alpha-glucose and C2 of beta-fructose. The bond formed between glucose and fructose is known as the 1,2-glycosidic bond. Also, sucrose is a dextrorotatory sugar, which changes to levorotatory, due to the dominant levorotatory nature of fructose. Therefore, the hydrolysis products of sucrose are D-glucose and D-fructose.

Isoelectric point refers to the \(\mathrm{H}^{+}\)ion concentration at become electrically neutral, the colloidal particles.

The\(\mathrm{H}^{+}\) concentration at which the colloidal particles have no charge is known as the isoelectric point. At this point, the stability of colloidal particles becomes very less, and they do not move under the influence of the electric field. The sol particles at the isoelectric point do not show the electrophoresis process.

In nitroprusside ion, the iron and NO exist as Fe(II) and \(NO^{+}\) rather than Fe(III) and NO. These forms can be distinguished by measuring the solid-state magnetic moment.

Nitroprusside ion is \(\left[\mathrm{Fe}(\mathrm{CN})_{5} \mathrm{NO}\right]^{2-}\). If the central atom iron is present here in \(\mathrm{Fe}^{2+}\) form, its effective atomic number will be \(26-2+(6 \times 2)=36\) and the distribution of electrons in valence orbitals (hybridised and unhybridized) of the \(\mathrm{Fe}^{2+}\) will be

It has no unpaired electron. So this anionic complex is diamagnetic. If the nitroprusside ion has \(\mathrm{Fe}^{3+}\) and \(\mathrm{NO}\), the electronic distribution will be such that it will have one unpaired electron i.e. the complex will be paramagnetic.

Thus, magnetic moment measurement establishes that in nitroprusside ion, the Fe and NO exist as F(II) and \(NO^{+}\) rather than Fe(III) and NO.

Boiling points of carbonyl compounds are higher than those of alkanes due to dipole-dipole interactions.

• The boiling points of aldehydes and ketones (carbonyl compounds) are higher than boiling point of non-polar alkanes due to the presence of dipole-dipole interactions between molecules of carbonyl compounds which are much stronger than Van der Waals forces between molecules of alkanes.
• So, energy required to break the interaction between carbonyl compound molecules is more than that of forces between molecules of alkane and therefore, boiling point is higher.

Trimethyl stearyl ammonium bromide is known as invert soap.

• Invert soap is a class of synthetic detergents in which the surface active part of the molecule is the cation.
• It is a cationic detergent. 
• Trimethyl stearyl ammonium bromide contains quaternary ammonium salt with an active cation part on the surface of the molecule. So, Trimethyl stearyl ammonium bromide is an invert soap.

The structure of Trimethyl stearyl ammonium bromide is as follows:

The addition polymers are formed by the repeated addition of monomer molecules possessing double or triple bonds, e.g., the formation of polythene from ethene and polypropene from propene. Addition polymers: Polyvinyl chloride, Polythene, Buna-S, Buna-N, etc. Polystyrene is a synthetic aromatic polymer made from monomer styrene. It is additional polymer as it is made of additional of monomers. The condensation polymers are formed by repeated condensation reactions between two different bi-functional or tri-functional monomeric units.

In these polymerization reactions, the elimination of small molecules such as water, alcohol, hydrogen chloride, etc. takes place. Examples are terylene (dacron), nylon 6, 6, nylon 6, etc. For example, nylon 6, 6 is formed by the condensation of hexamethylene diamine with adipic acid.

Gangue haveSand and soil,Silt andGravel.

The definition of gangue is worthless rock or mineral in which valuable minerals are found.Gangue are the impurities which are mixed closely with the ores. It is unwanted material which is of no market value and it is waste rock or mineral or sand etc.

On moving from left to right in a periodic table the electropositive character of elements decrease.

Electropositive character of an element is its ability to lose electrons and form positive ions. Now, as on moving from left to right in a period of periodic table, the nuclear charge increases due to the gradual increase in number of protons, so the valence electrons are pulled more strongly by the nucleus. Thus, it becomes more and more difficult for the atoms to lose electrons causing a decrease in the electropositive character of elements on moving from left to right in a periodic table.

2, 6-Dimethyldec-4-ene

Standard electrode potential of reaction will not change due to multiply the half-cell reactions with some numbers.

\(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}, E^{\circ}=-1.18 \mathrm{~V}\) ....... (i)

\(2\left(\mathrm{Mn}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}\right), E^{\circ}=+1.51 \mathrm{~V}\) ...... (ii)

Subtracting equation (ii) from equation (i), we get

\(3 \mathrm{Mn}^{2+} \longrightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}\)

\(E^{\circ}=-1.18-(+1.51)=-2.69 \mathrm{~V}\)

Since, the value of \(E^{\circ}\) is negative, therefore, the reaction is non-spontaneous.