Chemistry Test - 20

\(\begin{aligned} & \left(\Delta T_f\right)_x=\left(\Delta T_f\right)_y \\ & \mathrm{~K}_{\mathrm{f}} \mathrm{m}_{\mathrm{x}}=\mathrm{K}_{\mathrm{f}} \mathrm{m}_{\mathrm{y}} \\ & \mathrm{m}_{\mathrm{x}}=\mathrm{m}_{\mathrm{y}} \\ & \frac{\text { No. of moles of } \mathrm{x}}{\text { wt. of solvent in } \mathrm{kg}}=\frac{\text { No. of moles of } y}{\text { wt. of solvent in } \mathrm{kg}} \\ & \text { Given, } 4 \% \text { aqueous solution of } x=4 \mathrm{~g} \text { of solute, } x \text { present in } \\ & 100 \mathrm{~g} \text { of } \mathrm{H}_2 \mathrm{O} \\ & 12 \% \text { aqueous solution of } y=12 \mathrm{~g} \text { of solute, } y \text { present in } 100 \mathrm{~g} \\ & \text { of } \mathrm{H}_2 \mathrm{O} \\ & \text { Now, } \frac{4 g}{m_1} \times \frac{1000}{100}=\frac{12 g}{m_2} \times \frac{1000}{100} \\ & \frac{4 g}{\mathrm{~A}}=\frac{12 \mathrm{~g}}{\mathrm{~m}_2}\left[\mathrm{~m}_1=\mathrm{A}\right] \\ & \mathrm{m}_2=3 \mathrm{~A}\end{aligned}\)

Let, solubility of \(\mathrm{Ca}(\mathrm{OH})_2\) in pure water \(=\mathrm{Smol} / \mathrm{L}\)

\(\mathrm{Ca}(\mathrm{OH})_2 \rightleftharpoons \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-} \)

\( \mathrm{Smol} / \mathrm{L} 2 \times \mathrm{S}(\mathrm{mol} / \mathrm{L}) \)

\( =\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2=\mathrm{S} \times(2 \mathrm{~S})^2=4 \mathrm{~S}^3(\mathrm{~mol} / \mathrm{L})\)

The expression of \(\mathrm{K}_{\mathrm{sp}}\) can also be written as,

\( \mathrm{K}_{\mathrm{sp}}=\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}} \cdot \mathrm{S}^{\mathrm{x}+\mathrm{y}} \)

\(\quad=1^1 \cdot 2^2 \cdot \mathrm{S}^{1+2} \)

\( =4 \mathrm{~S}^3\)

\(\mathrm{K}_{\mathrm{sp}}=\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}} \cdot \mathrm{S}^{\mathrm{x}+\mathrm{y}} \)

\( \quad=1^1 \cdot 2^2 \cdot \mathrm{S}^{1+2} \)

\( \quad=4 \mathrm{~S}^3\left[\because \text { For } \mathrm{Ca}(\mathrm{OH})_2: \mathrm{x}=1, \mathrm{y}=2\right]\)

\(\mathrm{x}\) and \(\mathrm{y}\) are the coefficients of cations and anions respectively

\(S=\left(\frac{K_{s p}}{4}\right)^{1 / 3}=\left(\frac{5.5 \times 10^{-6}}{4}\right)^{1 / 3}\)

\( =1.11 \times 10^{-2} \mathrm{~mol} / \mathrm{L}\)

A bicycloalkene cannot have the formula of C₂H₆.

C₂H₆ has two degree of unsaturation (twoH₂less than saturated hydrocarbons), therefore it can be a diene, a cycloalkene or a bicycloalkane but it cannot be a bicycloalkene because it has three degrees of unsaturation.

\(\mathrm{HNO}_{3}\) pollutant is a component of Acid rains.

Acid rain: Acid rain is a by-product of a variety of human activities that emit the oxides of sulphur and nitrogen in the atmosphere.

Burning of fossil fuels (which contain sulphur and nitrogenous matter) such as coal and oil in power stations and furnaces or petrol and diesel in motor engines produce sulphur dioxide and nitrogen oxides. \(\mathrm{SO}_{2}\) and \(\mathrm{NO}_{2}\) after oxidation and reaction with water are major contributors to acid rain.

\(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \rightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\)

\(4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \rightarrow 4 \mathrm{HNO}_{3}(\mathrm{aq})\).

\(\begin{aligned} & \text { ( } 10 \mathrm{ml} \text { solute in } 90 \mathrm{ml} \text { solvent } \\ & \text { mass of solute }=10 \times 3.2=32 \mathrm{~g} \\ & \text { mass of solvent }=90 \times 1.6 \mathrm{~g} \\ & \mathrm{~m}=\frac{32 \times 1000}{160 \times 90 \times 1.6}=1.388 \\ & \mathrm{~m}=138.8 \times 10^{-2}=139\end{aligned}\)

Given, partial pressure of \(\mathrm{O}_2=20 \mathrm{kPa}\)

\(\mathrm{K}_{\mathrm{H}}\) (Henry's constant) \(=8 \times 10^4 \mathrm{kPa}\)

From Henry's law,

\(\mathrm{p}(\mathrm{g})=\left[\mathrm{K}_{\mathrm{H}}\right] \chi_{\mathrm{O}_2}\)

where, \(x_{\mathrm{O}_2}=\) solubility of oxygen

\(\begin{aligned}& 20 \times 10^3=\left(8 \times 10^4 \times 10^3\right) x_{0_2} \\& \Rightarrow x_{\mathrm{O}_2}=\frac{20}{8 \times 10^4}\end{aligned}\)

Solubility \(=2.5 \times 10^{-4}=25 \times 10^{-5}\)

The compounds that contain carbon are known as organic compounds.Most abundant organic compound on earth is cellulose.

Carbohydrates are grouped into monosaccharides, disaccharides and polysaccharides. Cellulose, a storage form of carbohydrate found in plants that humans cannot digest, is among the most plentiful of the carbohydrates worldwide.Cellulose is a polymer of linear polysaccharide, with several monosaccharide glucose units. The acetal linkage is beta which distinguishes it from starch.Human beings are unable to digest cellulose because there are no sufficient enzymes for breaking down beta-acetal linkages.

\(\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g})\)

Since \(\mathrm{H}_2\) is in excess and \(20 \mathrm{~L}\) of ammonia gas is produced.

Hence, 2 moles \(\mathrm{NH}_3 \equiv 1\) mole \(\mathrm{N}_2(\mathrm{v} \propto \mathrm{n})\)

\(20 \mathrm{LNH}_3 \equiv 10 \mathrm{LN}_2\)

Volume of \(\mathrm{N}_2\) left \(=56-10\)

\(=46 \mathrm{~L}\)

Let the solubility of \(\mathrm{A}_2 \mathrm{X}\) be ' \(\mathrm{S}\) '.

\(\begin{aligned}& \mathrm{A}_2 \mathrm{X}(\mathrm{s}) \rightleftharpoons 2 \mathrm{~A}^{+}(\mathrm{aq})+\mathrm{X}_{\mathrm{s}}^{2-}(\mathrm{aq}) ; \mathrm{K}_{\mathrm{sp} 1}=4 \times 10^{-12} \\& \mathrm{~K}_{\mathrm{sp}}=\left[\mathrm{A}^{+}\right]^2\left[\mathrm{X}^{2-}\right] \\& 4 \times 10^{-12}=(2 \mathrm{~S})^2(\mathrm{~S}) \\& 4 \times 10^{-12}=4 \mathrm{~S}^3 \\& \Rightarrow \mathrm{S}=10^{-3} \mathrm{M}\end{aligned}\)

Let the solubility of \(\mathrm{MY}\) be ' \(\mathrm{S}_1\) '.

\(\begin{aligned}& \mathrm{MY}(\mathrm{s}) \rightleftharpoons \mathrm{M}^{+}(\mathrm{aq})+\mathrm{Y}^{-}(\mathrm{aq}), \mathrm{K}_{\mathrm{sp}_2}=4 \times 10^{-12} \\& \mathrm{~K}_{\mathrm{sp}_2}=\left[\mathrm{M}^{+}\right]\left[\mathrm{Y}^{-}\right] \\& \mathrm{K}_{\mathrm{sp}_2}=\left(\mathrm{S}_1\right)^2 \\& 4 \times 10^{-12}=\left(\mathrm{S}_1\right)^2 \\& \Rightarrow \mathrm{S}_1=2 \times 10^{-6} \mathrm{M} \\& \frac{\left[\mathrm{A}_2 \mathrm{Y}\right]}{[\mathrm{MY}]}=\frac{\mathrm{S}}{\mathrm{S}_1}=\frac{10^{-4}}{2 \times 10^{-6}}=50 \\&\end{aligned}\)

Benzyl chloride gives benzoic acid on oxidation.

The reaction takes place as follows:

\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Cl}+2 \mathrm{KOH}+2[\mathrm{O}] \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOK}+\mathrm{KCl}+\mathrm{H}_{2} \mathrm{O}\)