Chemistry Test - 29

\(\begin{aligned} & \mathrm{NH}_3+\mathrm{HCl} \rightarrow \mathrm{NH}_4 \mathrm{Cl} \\ & \text { moles of } \mathrm{HCl}=0.2 \mathrm{M} \times 25 \times 10^{-3} \mathrm{~L}=0.005 \text { moles } \mathrm{HCl} \text { (total } \\ & \text { consumed) } \\ & \text { moles of } \mathrm{NH}_3=0.2 \mathrm{M} \times 50 \times 10^{-3} \mathrm{~L}=0.01 \mathrm{~mol} \mathrm{esH} \mathrm{Cl} \\ & \text { excess } \mathrm{NH}_3=0.01-0.005=0.005 \mathrm{~mol} \text { es } \\ & 1 \text { mole ammonia }=1 \mathrm{~mol} \mathrm{eN} \mathrm{H}_4 \mathrm{Cl} \\ & 0.005 \mathrm{NH}_3=0.005 \mathrm{NH}_4 \mathrm{Cl} \\ & \text { Total volume }=\mathrm{V}_{\mathrm{HCl}}+\mathrm{V}_{\mathrm{NH}_3}=25+50=75 \mathrm{~mL} \\ & {\left[\mathrm{NH}_3\right]=\left[\mathrm{NH}_4 \mathrm{Cl}\right]=\frac{0.005 \mathrm{molel}^{-3}}{75 \times 10^{-3} \mathrm{~L}}=0.066 \mathrm{M}} \\ & \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{\left[\mathrm{NH}_4 \mathrm{Cl}\right]}{\left[\mathrm{NH}_3\right]} \\ & \mathrm{pOH}=4.75+\log \frac{[0.066]}{[0.066]} \\ & \mathrm{pOH}=4.75 \\ & \mathrm{pH}=14-\mathrm{pOH} \Rightarrow \mathrm{pH}=9.25\end{aligned}\)

Formaldehyde is also known as formalin.

It is an organic compound which occurs naturally with formula CH₂O (H−CHO).

It is an effective precursor for many other chemicals and compounds.

This is primarily used in industrial resin processing, e.g. for particle boards and coatings.

Acetaldehyde and Acrolein are the Aldehydes while Acetone is the Ketones.

There is a general trend that half filled and completely filled orbitals are more stable than incompletely filled orbitals.

\(\mathrm{C u(II)}\) is more stable than \(\mathrm{C u(I)}\) because it has a \(2^{+}\) charge and is smaller than \(\mathrm{C u(I)}\).

The electron density and effective nuclear charge is greater for \(\mathrm{C u(II)}\) and it forms stronger bonds (high hydration energy) and releases more energy and is more stable.

Therefore, \(\mathrm{C u(I I)}\) is more stable than \(\mathrm{C u(I)}\).

\(\mathrm{NaHCO}_{3}\) is amphiprotic species just like \(\mathrm{NaHSO}_{4}, \mathrm{NaHS}, \mathrm{Na}_{2} \mathrm{HPO}_{4}, \mathrm{NaH}_{2} \mathrm{PO}_{4}\) etc.

In such cases the solution \(\mathrm{pH}\) is dependent on the two ionization constants of the amphiprotic substance and independent of the concentration of the amphiprotic substance in the solution. However, they do not hold pH well and are not effective as buffer solutions. Here the trick is to use the formula,

\(\mathrm{pH}=\frac{1}{2}\left(\mathrm{pK}_{\mathrm{at}}^{0}+\mathrm{pK}_{\mathrm{a} 2}^{0}\right)\) where \(\mathrm{K}_{\mathrm{a} 1} \& \mathrm{~K}_{\mathrm{a} 2}\) are dissociation constants of the conjugate acid of amphiprotic species.

For \(\mathrm{NaHCO}_{3}, \mathrm{pH}\) is \(\frac{(6.37+10.25)} {2}=8.31\)

\(40 \mathrm{~g}\) of glucose mixed with \(200 \mathrm{~mL}\) of water

\(180 \mathrm{~g}\) of glucose \(=1\) moles of glucose

\(40 \mathrm{~g}\) of glucose \(\mathrm{mol}=0.22 \mathrm{~mol}\)

\(1 \mathrm{~mL}\) of water \(=1 \mathrm{~g}\) of water \(\left[\mathrm{d}=1 \mathrm{~g} / \mathrm{cm}^2, 1 \mathrm{~mL}=1 \mathrm{~cm}^3\right]\)

\(200 \mathrm{~mL}\) of water \(=200 \mathrm{~g}\) of water

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

where, \(\Delta \mathrm{T}_{\mathrm{f}}=\) depression in freezing point,

\(\mathrm{K}_{\mathrm{f}}=\) molal elevation constant \(=1.86 \mathrm{Kkg} \mathrm{mol}^{-1}\) and

\(\mathrm{m}=\) molality of solution

\(\Delta \mathrm{T}_{\mathrm{f}}=\frac{1.86 \times 0.22}{200} \times 1000,(\because 1000 \mathrm{~g}=1 \mathrm{~kg}) \)

\(\Delta \mathrm{T}_{\mathrm{f}}=2 \mathrm{~K} \)\

\(\therefore\left(T_f-T_f^{\prime}\right)=2\left[\begin{array}{c}T_f=\text { freezing point of water }(273 \mathrm{~K}) \\ T_f^{\prime}=\text { freezing point of solution }\end{array}\right]\)

\(273 \mathrm{~K}-\mathrm{T}_{\mathrm{f}}^{\prime}=2 \mathrm{~K} \)

\(\mathrm{~T}_{\mathrm{f}}^{\prime}=(273-2) \mathrm{K}=271 \mathrm{~K}\)

Given:\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCl}\)

Numbering the carbon atoms as:

\(\mathrm{\stackrel{3}{C}H_{3}\stackrel{2}{C}H_{2}\stackrel{1}{C}OCl}\)

• This is an acid chloride.
• It consists of three carbon atoms in the parent chain and chloroformyl group \(\mathrm{(-COCl)}\).
• In the IUPAC name, the suffix "oyl chloride" is added.Therefore, the IUPAC name will be propanoyl chloride.

The catalytic efficiency of an enzyme is described by the \(\mathrm{K}_{\mathrm{m}}\) value or Michaelis Menten constant. The Michaelis constant is the substrate concentration at which the reaction rate is one half the maximum. The \(\mathrm{K}_{\mathrm{m}}\) describes the affinity of enzyme for a substrate molecule. Greater the affinity lower is the \(\mathrm{K}_{m}\) value and sooner the \(\mathrm{V}_{\max }\) can be attained.

According to Michaelis-Menten equation \(\mathrm{K}_{\mathrm{m}}\) is equal to substrate concentration at which the velocity is half the maximum. Michaelis and Menten proposed a hypothesis for enzyme for action according to which the enzyme molecule combines with a substrate complex which further dissociates to form product and enzyme back.

First Step :

\(\mathrm{MnO}_2+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_2+\mathrm{Cl}_2+2 \mathrm{H}_2 \mathrm{O}\)

Here \(1 \mathrm{~mol}\) of \(\mathrm{MnO}_2\) produce \(1 \mathrm{~mol}\) of \(\mathrm{Cl}_2\)

\(\therefore\) Mole ratio of \(\mathrm{n}_{\mathrm{M}_{\mathrm{nO}_2}}: \mathrm{n}_{\mathrm{Cl}_2}=1: 1\)

Second Step :

\(\mathrm{Cl}_2+2 \mathrm{Kl} \rightarrow 2 \mathrm{KCl}+\mathrm{I}_2\)

Here, \(1 \mathrm{~mol}\) of \(\mathrm{Cl}_2\) produce \(1 \mathrm{~mol}\) of \(\mathrm{I}_2\)

Mole ratio of \(\mathrm{n}_{\mathrm{Cl}_2}: \mathrm{n}_{\mathrm{I}_2}=1: 1\)

Third Step :

\(\mathrm{I}_2+2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3 \rightarrow 2 \mathrm{Nal}+\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3\)

\(1 \mathrm{~mol}\) of \(\mathrm{I}_2\) react with \(2 \mathrm{~mol}\) of \(\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3\)

Mole ratio of \(\mathrm{n}_{\mathrm{I}_2}: \mathrm{n}_{\mathrm{N}_2 \mathrm{~S}_2 \mathrm{O}_3}=1: 2\)

Given \(\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3\) is \(60 \mathrm{~mL}\) of \(0.1 \mathrm{M}\)

\(\therefore\) Number of moles of \(\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3\)

\(=\mathrm{V}(\) in \(\mathrm{L}) \times \mathrm{M}\) (Molarity)

\(=\frac{60}{1000} \times 0.1\)

\(=0.006 \mathrm{~mol}\)

\(\therefore\) Number of moles of \(\mathrm{I}_2\)

\(=\frac{1}{2}(0.006) \)

\(=0.003\)

\(\therefore\) Moles of \(\mathrm{MnO}_2=0.003\) (as mole ratio of \(\mathrm{MnO}_2\) and \(\mathrm{Cl}_2=1: 1\) )

Molar mass of \(\mathrm{MnO}_2=55+32=87\)

\(\therefore\) Mass of \(\mathrm{MnO}_2=0.003 \times 87=0.261 \mathrm{gm}\)

Given \(\mathrm{MnO}_2=2 \mathrm{~g}\)

\(\therefore \%\) of \(\mathrm{MnO}_2=\frac{0.261}{2} \times 100=13 \%\)

In ionic solids the vacancies are produced due to absence of cations and anions in stoichiometric proportions resulting in defect called Schottky defect.

The density is less than expected due to missing spaces.

The bond between carbon and the halogen atom \((\mathrm{C}-\mathrm{X})\) is polar.

• Since halogen atoms are more electronegative than carbon, there is partial negative charge on halogen atom and positive charge on carbon.
• This makes the molecule dipolar and polar nature of \((\mathrm{C}-\mathrm{X})\) bond increases with an increase in the electronegativity difference between carbon and halogen atom.

Therefore, the order of polarity of \(\mathrm{C}-\mathrm{X}\) bond will be:

\(\mathrm{C}-\mathrm{F}>\mathrm{C}-\mathrm{Cl}>\mathrm{C}-\mathrm{Br}>\mathrm{C}-\mathrm{I}\)

Thus, the increasing order of electronegativity of halides will be:

\(\mathrm{I, B r, Cl, F}\)