Chemistry Test - 7

Mosley gave the modern Long form of Periodic table where the base of classification was the atomic number.

According to Mosley's periodic law, the physical and chemical properties of elements are periodic functions of their Atomic number.

In the long-form periodic table, horizontal rows are called Periods and vertical rows are called groups.

Group 13 contains Aluminium (13) and Gallium (31).

Element with atomic number 12 is magnesium which belongs to group 2 and element with atomic number 30 is zinc which belongs to group 12.

Element with atomic number 11 is sodium which belongs to group 1 and element with atomic number 20 is calcium which belongs to group 2.

Element with atomic number 14 is silicon belonging to group 14 and 31 is Gallium from group 13.

So, 13, 31 pairs of atomic numbers represent elements belonging to the same group (Group 13).

Number of atoms in \(20 \mathrm{~g}\) \(\mathrm{Ca}=\frac{\text { weight }}{\text { Atomic weight }} \times \mathrm{N}_{\mathrm{A}}\)

\(=\frac{20}{40} \times 6.023 \times 10^{23}\)\(=3.0115 \times 10^{23}\) atoms

(A) Number of atoms in \(24 \mathrm{~g}\) \(\mathrm{Mg}=\frac{24}{24} \times 6.023 \times 10^{23}=6.023 \times 10^{23}\) atoms

(B) Number of atoms in \(8 \mathrm{~g}\) Oxygen \(=\frac{8}{16} \times 6.023 \times 10^{23}=3.0115\) \(\times 10^{23}\) atoms

(C) Number of atoms in \(12 \mathrm{~g}\) Carbon \(=\frac{12}{12} \times 6.023 \times 10^{23}=6.023\)\(\times 10^{23}\) atoms

(D) Number of atoms in \(16 \mathrm{~g}\) Oxygen \(=\frac{16}{16} \times 6.023 \times 10^{23}=6.023\)\(\times 10^{23}\) atoms

P is true while Q, R and S are false

The stereochemistry of steroidal aglycones in cardiac glycosides rings A-B and C-D are cis fused while B-C is transfused.Stereochemistry, a subdiscipline of chemistry, involves the study of the relative spatial arrangement of atoms that form the structure of molecules and their manipulation.

Important Structural Element of Cardiac Glycosides:steroid portion of cardiac glycoside molecule has cis fused A/B and C/D rings (B/C is trans fused) -fused ring system forms a “U” shape -the activity of the steroid is dependent on that specific conformation -hydroxyl groups affect the distribution and affect duration of action.

When methyl Iodide is treated with sodium in the presence of dry ether, Ethane is formed.

It iscalled Wurtz reaction.

\(\mathrm{CH}_{3}-\mathrm{I}+2 \mathrm{Na}+\mathrm{I}-\mathrm{CH}_{3} \overset{_\text{Dry ether}}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{CH}_{3}+2 \mathrm{NaI}\)

Some of the polar crystals when heated produce an electric current. This phenomenon is called Pyro-electricity or pyroelectric effect.

Pyroelectricity is a property of certain crystals which are naturally electrically polarized and as a result contain large electric fields. Pyroelectricity can be described as the ability of certain materials to generate a temporary voltage when they are heated or cooled.

According to the Pauli exclusion principle no two electrons can have the same set of all four quantum numbers simultaneously. Two electrons present in the same orbital can be distinguished by their spin quantum number. If one of them has a \(+1/2\) value then other must have the \(-1/2\) value.

Option (A) is not correct.

Classical smog is the smog that occurs in cold and humid climates and it contains compounds like sulphur dioxide, particulate matter that are compounds of reducing nature. Option (A) is the definition of photochemical smog.

The pair having the same magnetic moment is \(\left[ Cr \left( H _{2} O \right)_{6}\right]^{2+}\) and \(\left[ Fe \left( H _{2} O \right)_{6}\right]^{2+}\).

In \(\left[ Cr \left( H _{2} O \right)_{6}\right]^{2+}\), the chromium metal is in \(+2\) oxidation state with \(3 d ^{4}\) outer electronic configuration and 4 unpaired electrons.

In \(\left[ Fe \left( H _{2} O \right)_{6}\right]^{2+}\), the iron metal is in \(+2\) oxidation state with \(3 d ^{6}\) outer electronic configuration and 4 unpaired electrons.

Since both complexes have same number of unpaired electrons, they have the same magnetic moment.

In the question it is given that the concentration of \(\mathrm{NH}_{4} \mathrm{Cl}\) is \(0.02 \mathrm{M}\) and base dissociation constant of \(\mathrm{NH}_{4} \mathrm{OH} \text { is } 10^{-5}\).

We have to find the \(\mathrm{pH}\) of the \(0.02 \mathrm{M} \mathrm{NH} 4 \mathrm{Cl}\) solution from the given data.

To calculate the pH of the \(\mathrm{NH}_{4} \mathrm{Cl}\) solution (formed by strong acid and weak base) there is a formula. The formula is as follows.

\(p H=7-\frac{1}{2}\left[p K_{b}+\log C\right]\)

Here pH = pH of the salt solution

\(K_{b}=\) base dissociation constant \(\left(p K_{b}=-\log k_{b}\right)\)

C = concentration of the salt

Now we have to substitute all the known values in the above formula to get the pH of the solution.

\(p H=7-\frac{1}{2}\left[p K_{b}+\log C\right]\)

\(=7-\frac{1}{2}\left[\log 10^{-5}+\log 0.02\right]\)

\(=7-\frac{5}{2}-\frac{1}{2}\left(\log 2 \times 10^{-2}\right)\)

\(=5.35\)

Therefore the pH of the ammonium chloride solution is 5.35.