Chemistry Test - 23

Number of atoms in \(1 {~g}\) of gold \(=\frac{\text { Avogadro's No. of atoms }}{\text { Gram atom of gold }}\)

\(=\frac{\left(6.022 \times 10^{23} {~mol}^{-1}\right)}{\left(197 {~g} {~mol}^{-1}\right)} \times(1 g)\)

Number of atoms per unit cell (fcc) of gold \(=4\)

\(\therefore\) Number of unit cells in \(1 {~g}\) of gold \(=\frac{\left(6.022 \times 10^{23} {~mol}^{-1}\right) \times(1 {~g}) \times 1}{\left(197 {~g} {~mol}^{-1}\right) \times 4}\)

\(=7.64 \times 10^{20}\)

Different gases have different \(\mathrm{K}_{\mathrm{H}}\) values at the same temperature. Thissuggests that \(\mathrm{K}_{\mathrm{H}}\) is a function of the nature of the gas.

\(\therefore\) Statement (A) and (D) both are correct.

The most commonly used form of Henry's law states that "the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution" and is expressed as:

\(\mathrm{p}=\mathrm{K}_{\mathrm{H}} \mathrm{X}\)

Here \(\mathrm{K}_{\mathrm{H}}\) is Henry's law constant.

Given that:

\({T}=673 {~K}, {~K}_{{c}}=0.5,\)

We know that:

\({R}=0.082\) litre bar \({K}^{-1} {~mol}^{-1}\)

\(K_{p}=K_{c}[R T]^{n}\)

Where,

\(n=\) Concentration of products - concentration of reactants

\(=2-4\)

\(n=-2\)

Then,

\({K}_{{p}}=0.5 \times(0.082 \times 673)^{-2}\)

\(=1.64 \times 10^{-4} {~atm}\)

Let 1 and 2 denote solvent and solute respectively,

\(\mathrm{x}_2=\frac{\mathrm{n}_2}{\mathrm{n}_1+\mathrm{n}_2}=\frac{\mathrm{n}_2}{\frac{\mathrm{m}_1}{\mathrm{M}_1}+\mathrm{n}_2}\)

\(=\frac{\mathrm{n}_2 \mathrm{M}_1}{\mathrm{~m}_1+\mathrm{n}_2 \mathrm{M}_1}=\frac{\mathrm{n}_2 \mathrm{M}_1}{\left(\mathrm{~m}_1+\mathrm{m}_2\right)+\mathrm{n}_2 \mathrm{M}_1-\mathrm{m}_2}\)

\(=\frac{\mathrm{n}_2 \mathrm{M}_1}{\mathrm{~V}_{\mathrm{e}}+\mathrm{n}_2 \mathrm{M}_1-\mathrm{m}_2}=\frac{\mathrm{n}_2 \mathrm{M}_1}{\mathrm{~V}_{\mathrm{p}}+\mathrm{n}_2\left(\mathrm{M}_1-\mathrm{M}_2\right)}\)

\(=\frac{\left(\frac{\mathrm{n}_2}{\mathrm{v}}\right) \mathrm{M}_1}{\mathrm{p}+\left(\frac{\mathrm{n}_2}{\mathrm{v}}\right)\left(\mathrm{M}_1-\mathrm{M}_2\right)}=\frac{\mathrm{MM}_1}{\mathrm{p}+\mathrm{M}\left(\mathrm{M}_1-\mathrm{M}_2\right)}\)

The compound which has one isopropyl group is2- Methypentane.

In the mixture of nitrite and nitrate ions, the nitrite ions can be removed by heating with urea and supluric acid.

The reaction takes place as follows:

\(2 \mathrm{NaNO}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{HNO}_{2}\)

\(2 \mathrm{HNO}_{2}+\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{N}_{2}+3 \mathrm{H}_{2} \mathrm{O}\)

In the above reaction, we can see that the products, nitrogen and carbon dioxide are in gaseous form and thus, do not stay in the solution. This reaction is exothermic and thus, will occur spontaneously without the supply of any extra energy.

Given that:

Concentration of \({Na}_{2} {CO}_{3}=1.0 \times 10^{-4} {M}\)

\(\therefore\left[{CO}_{3}^{-}\right]=1.0 \times 10^{-4} {M}\)

i.e., \(s=1.0 \times 10^{-4} {M}\)

At equilibrium:

\(\left[{Ba}^{++}\right]\left[{CO}_{3}^{--}\right]={K}_{{sp}}\) of \({BaCO}_{3}\)

\(\left[{Ba}^{++}\right]=\frac{K_{s p}}{\left[{CO}_{3}^{--}\right]}\)

\(=\frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}}\)

\(=5.1 \times 10^{-5} {M}\)

According to the ideal gas equation,

\(\mathrm{PV}=\mathrm{nRT}\)

\(\mathrm{n}=\frac{\text { Mass given }}{\text { Molecular mass of } \mathrm{O}_{2}}\)

\(=\frac{5}{32}\)

\(\mathrm{PV}=\left(\frac{5}{32}\right) \mathrm{RT}\)

The compound having the least molecular mass should contain the minimum amount of sulphur or simply 1 atom of sulphur.

Let the molecular mass of the compound (in amu) be \(x\).

Mass of sulphur in the compond \(=8\%\) of the total molecular mass of \(x\)

\(=\frac{8 x}{100}\)

Since, the molecular mass of one mole of sulphur is 32, the compound must contain this amount of sulphur, i.e.,

\(\frac{8 x}{100}=32\)

which gives:

\(x=400\)

Allylic halides are the halides in which the halogen atom is bonded to \(sp^3-\) hybridized carbon atom next to carbon-carbon double bond \((C=C)\).

General chemical structure of allylic halides:

\({CH}_2={CH}-{CH}_2-{X}\)

Here, \(X\) is any halogen atom.

So, according to the rule, only in \(3-\) chloro cyclohex\(-1-\)ene halide \((Cl)\) is attached with \(sp^3\) carbon which is next to the carbon-carbon double bond.