Chemistry Test - 28

Given:

Weight of solute \(({w})=1 {~g}\)

Weight of solvent \(({W})=75 {~g}\)

Boiling point of solution \(=100.114^{\circ} {C}\)

Boiling point of solvent \(=100^{\circ} {C}\)

Therefore,

\(\triangle {T}=100.114-100=0.114^{\circ} {C}\)

Molecular weight of solute \(({m})=60.1\)

Boiling point elevation constant \(({K})=?\)

We know that:

\({m}=\frac{1000 \times {K} \times {w}}{\Delta {T} \times {W}}\)

or, \({K}=\frac{{m} \times \Delta {T} \times {W}}{1000 \times {w}}\)

\(=\frac{60.1 \times 0.114 \times 75}{1000 \times 1}\)

\(=\frac{513.8}{1000}\)

\(K=0.513\)

In physisorption adsorbent does not show specificity for any particular gas because involved Van der Waals forces are universal.

This phenomenon involves the use of weak Van der Waal forces by means of which gas molecules get adsorbed on a solid surface.There is no specificity as any gas can be adsorbed onto the surface.

We know that for a first order reaction:-

Rate = k[R]

\(\therefore 1.5 \times 10^{-2} \mathrm{M} \min ^{-1}=\mathrm{k}[0.5 \mathrm{M}]\)

\(\Rightarrow \mathrm{k}=\frac{1.5 \times 10^{-2} \mathrm{M} / \mathrm{min}}{0.5 \mathrm{M}}\)

\(\Rightarrow \mathrm{k}=0.03 \mathrm{~min}^{-1}\)

We also know that, half life of a first order reaction:

\(\mathrm{t} \frac{1 } 2=\frac{ln (2)}{\mathrm{k}}\)

\(=\frac{0.693}{\mathrm{k}}\)

\(=\frac{0.693}{0.03 \mathrm{~min}^{-1}}\)

\(=23.1 \) min

An aqueous solution of hydrochloric acidshows negative deviation from Raoult's law.

Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present, i.e.,

\(\left[\mathrm{P}_{\text {solution }}=\mathrm{X}_{\text {solvent }} \mathrm{P}_{\text {solvent }}^{\circ}\right]\)

If the attraction between different molecules, for example between \(\mathrm{HCl}\) and \(\mathrm{H}_{2} \mathrm{O}\) molecules, is stronger, the escaping tendency from the solution to the vapour phase will be smaller, then the partial vapour pressure will be smaller than predicted by Raoult's law and the system exhibits a negative deviation.

Unsaturated hydrocarbons are hydrocarbons that have double or triple covalent bonds between adjacent carbon atoms.

The term "unsaturated" means more hydrogen atoms may be added to the hydrocarbon to make it saturated (i.e. consisting of all single bonds).

The example of the unsaturated compound is alkenes and alkynes, as well as branched chains and aromatic compounds.

SO₂:

In Sulphur dioxide, the dipole actas from sulphur towards oxygen and from sulphur towards the lone pair of electrons.

The net dipole is acting downwards and the net dipole is thus not zero.

Beryllium carbonate (\(BeCO _{3}\))is unstable due to the smaller size of cation and larger size of the anion (as smaller cation stabilizes smaller anion through crystal lattice energy) and can be kept only in the atmosphere of \(CO _{2}\). The thermal stability increases with increasing cationic size. So, \(BeCO _{3}\) is unstable in the air.

The percentage of s- character of the hybrid orbitals in ethane, ethene and ethyne are respectively 25, 33, 50.

Calgon is used to remove permanent hardness of water by removing cation \(\mathrm{Mg}^{2+}\).

Calgon is not an element, it is a compound. The \(2 \mathrm{nd}\) most abundant element by weight in the Earth's crust is silicon (Si). So, the statement in option (a) is not true.

Calgon is sodium hexametaphosphate \(\left(\mathrm{NaPO}_3\right)_6\) or \(\mathrm{Na}_2\left[\mathrm{Na}_4\left(\mathrm{PO}_3\right)_6\right]\). It is a polymeric water soluble compound. It is also named as Graham's salt.

\(\mathrm{Mg}^{2+}\) of hard water is removed as

\(\underset{\text{Calgon}}{\mathrm{Na}_2\left[\mathrm{Na}_4\left(\mathrm{PO}_4\right)_3\right]}+2 \mathrm{M} \mathrm{g}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Na}_2\left[\mathrm{Mg}_2\left(\mathrm{PO}_3\right)_6\right](\mathrm{s})+4 \mathrm{Na}^{+}(\mathrm{aq})\)

\(\mathrm{Ca}^{2+}\) ion cannot be removed by calgon as \(\mathrm{Ca}^{2+}\) is larger in size than \(\mathrm{Mg}^{2+}\).

So, the statement of option (b), (c) and (d) are true.

\(\mathrm{C}_{2} \mathrm{H}_{6}\) compound has \(7\) covalent bonds.

The structure of the Ethane molecule is given below.

From the above structure, we can see that the Ethane molecule has \(6~ \mathrm{C}-\mathrm{H}\) covalent bonds and one \(\mathrm{C-C}\) covalent bond. Therefore, on adding the covalent bonds of \(\mathrm{C}-\mathrm{C}\) and \(\mathrm{C}-\mathrm{H}\) covalent bonds, we can say that the Ethane molecule has \(7\) covalent bonds.