Chemistry Test - 5

For solute \(S\)
\(3 S \rightleftharpoons S _{3}\) (Initially)
1 0
1-a \(\quad \frac{a}{3}\) (at equilibrium)
so vant hoff factor for solute s, \(i_{1}=1-\alpha+\frac{\alpha}{3}=1-\frac{2 \alpha}{3}\)
for other solute \(i_{2}=1\) (as neither assosiacte nor dissociate) Now elevation of boiling point in both solution is same so molality of first solvent \(x\) vant hoff factor for first solvent \(=\) molality of second solvent \(x\) vant hoff factor for second solvent \(\Rightarrow 0.1 i_{1}=0.08 i_{2}\)
\(\Rightarrow 0.1\left(1-\frac{2 n}{3}\right)=0.08\)
\(\Rightarrow \alpha=0.3\)
\(\Rightarrow 30 \%\) trimerization

Relative humidity \(( RH )=\frac{\text { Partial pressure of } H _{2} O \text { in air }}{\text { Vapour pressure of } H _{2} O }\)
Partial pressure of \(H _{2} O = RH \times\) Vapour pressure of \(H _{2} O\) \(=\frac{40}{100} \times 17.5=7\) torr
\(=\frac{7}{760} atm =0.0092 atm\)
Now \(PV = nRT\)
\(n=\frac{P V}{R T}=\frac{0.0092 \times 1}{0.082 \times 293}\)
\(=0.000382\) mole
\(=3.82 \times 10^{-4} mole\)

Aromaticity of a compound can be decided by Huckel’s rule.
Huckel’s rules are as follows:
1. Compound must be cyclic
2. The cyclic compound must have resonance and complete delocalization of electrons i.e., there must be conjugation.
3. The total number of pi electrons must fit the formula = 4n + 2, where n is a non – negative integer.
The important summary of this concept is as follows: (Remember this summary)
\begin{equation}\begin{array}{|c|c|c|}
\hline \text { Aromatic Compound } & \text { Anti - Aromatic Compound } & \begin{array}{c}
\text { Non - aromatic } \\
\text { Compound }
\end{array} \\
\hline \text { Cyclic } & \text { Cyclic } & - \\
\hline \text { Conjugation of electrons } & \text { Conjugation of Electrons } & - \\
\hline(4 n +2) \text { pi electrons } & \text { 4n electrons } & - \\
\hline \text { Flat Structure } & \text { Flat Structure } & - \\
\hline
\end{array}\end{equation}(a)
\begin{equation}\Rightarrow 4 n+2=2\end{equation}
\(\Rightarrow \underline{n}=0\) (aromatic)
(b)
\begin{equation}\Rightarrow 4 n+2=6\end{equation}
\(n =1(\) aromatic \()\)
(c)
\begin{equation}\Rightarrow 4 n+2=4\end{equation}
\(n =\frac{1}{2}( n\) is not integer \()\) hence not aromatic
(d)
\begin{equation}\Rightarrow 4 n+2=6\end{equation}
N=1(aromatic)

This test is meant for detecting the coloured cations. When borax is heated on a platinum loop it swells up into a white porous mass first which on subsequent heating melts into a shining transparent glass bead (B2O3).

(i) \(K _{3}\left[ Fe ( CN )_{6}\right]\)
\(Fe ^{+3}\) has \(3 d ^{5}\) configuration, \(CN ^{-}\) is strong field ligand and thus \(Fe ^{+3}\) has four electron paired leaving one unpaired electron with \(d ^{2} sp ^{3}\) hybridization and thus paramagnetic.
(ii) \(\left[ Co \left( NH _{3}\right)_{6}\right] Cl _{3}\)
\(Co ^{3+}\) has \(3 d ^{6}\) configuration, \(NH _{3}\) is strong field ligands thus all six electrons are paired with \(d ^{2} sp ^{3}\) hybridization and therefore diamagnetic.
(iii) \(K _{2}\left[ Pt ( CN )_{4}\right]\)
\(Pt ^{+2}\) has \(3 d ^{8}\) configuration, \(CN ^{-}\) is strong field ligand and thus all the eight electrons are paired with dsp \(^{2}\) hybridization and therefore diamagnetic.
(iv) \(\left[ Zn \left( H _{2} O \right)_{6}\right]\left( NO _{3}\right)_{2}\)
\(Zn ^{+2}\) has \(3 d ^{10}\) configuration and thus all paired, showing \(sp ^{3} d ^{2}\) hybridization and so diamagnetic.

Let solubility of AgCl in water =s

\(k _{ sp } AgCl = s ^{2}\)
\(s^{2}=1.8 \times 10^{-10}\)
Now in \(0.01 M KCl\) aqueous solution
Complete hydrolysis of \(KCl\) will take place and there will be a concentration of \(\left[ Cl ^{-}\right]=0.01 M\) \(0.01 MKCl \longrightarrow 0.01 MK ^{+}+0.01 MCI ^{-}\)
Now if solubility of \(AgCl\) in \(KCl\) solution is \(s ^{\prime}\)
Then

\(k _{ sp } Ag Cl = s ^{\prime}\left(0.01+ s ^{\prime}\right)\)
\(s ^{\prime}\) is small compare to 0.01 so neglecting it with respect to 0.01 \(k _{ sp } AgCl = s ^{\prime}(0.01)\)
\(1.8 \times 10^{-10}= s ^{\prime} \times 10^{-2}\)
\(s^{\prime}=1.8 \times 10^{-8} mol L ^{-1}\)
= 1.8 × 10^-8 mol dm^-3 (1dm³= 1 Liter)

‘N’ is in second period so it cannot expand its octet, hence it can only accommodate 8 electrons in the outer shall so it can only form NCl₃ not NCl₅ which has 10 electrons, while P can expand its octet so it can form both Pcl₃ and Pcl₅.

Hence option A is correct.

Acidic character depends on stability of conjugate base. More stable conjugate base, more will be the acidic character

Due to the small size of oxide ion than sulphide ion, Metal oxides are generally less stable.

Hence option D is correct.

Paracetamol, also known as acetaminophen, is a medication used to treat pain and fever. It is typically used for mild to moderate pain relief. Evidence is mixed for its use to relieve fever in children. It is often sold in combination with other medications, such as in many cold medications. 

In addition to antimalarial medication, millions of children with Plasmodium falciparum malaria receive paracetamol to reduce fever. However, the usefulness of this practice has not been proven.

Paracetamol is termed a simple analgesic and an antipyretic. Despite enduring assertions that it acts by inhibition of cyclooxygenase (COX)-mediated production of prostaglandins, unlike non-steroidal anti-inflammatory drugs (NSAIDs), paracetamol has been demonstrated not to reduce tissue inflammation.

Hence option D is correct.