Chemistry Test - 3

According to Freundlich isotherm,

\(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kp}^{\frac{1}{n}} \)

\((\text { Using, amount of adsorbate ∝ Volume of absorbate) } \)

\(\frac{10}{1}=\mathrm{k} \times(100)^{\frac{1}{n}} ....\text{(i)} \)

\(\frac{15}{1}=\mathrm{k} \times(200)^{\frac{1}{\mathrm{n}}} ....\text{(ii)}\)

\(\frac{\mathrm{v}}{1}=\mathrm{k} \times(300)^{\frac{1}{n}}  ....\text{(iii)}\)

Divide Eq. (ii) by (i)

\(\frac{15}{10}=2^{\frac{1}{x}} \)

\(\Rightarrow \log \left(\frac{3}{2}\right)=\frac{1}{n} \log 2 \)

\(\frac{1}{n}=\frac{\log 3-\log 2}{\log 2}=\frac{0.4771-0.3010}{0.3010} \)

\(\frac{1}{n}=0.585\)

Divide Eq. (iii) by (i)

\(\frac{\mathrm{v}}{10}=3^{\frac{1}{n}}\)

\(\log \left(\frac{\mathrm{v}}{10}\right)=\frac{1}{\mathrm{n}} \log 3\)

\(\log \left(\frac{\mathrm{v}}{10}\right)=0.585 \times 0.4771=0.2791 \)

\(\frac{\mathrm{v}}{10}=10^{0.2791} \)

\(\Rightarrow \mathrm{v}=10 \times 10^{0.2791} \)

\(=10^{1.279}=10^{\mathrm{x}} \)

\(\mathrm{x}=1.279 \)

\(\mathrm{x}=128 \times 10^{-2}\)

p-aminophenol is formed on the electrolytic reduction of nitrobenzene in presence of strong acid.

The electrolytic reduction of nitrobenzene in strongly acidic medium produces phenylhydroxylamine which rearranges to p-Aminophenol.

In weakly acidic medium, aniline is obtained whereas in alkaline medium, various mono and di-nuclear reduction products (such as nitrosobenzene, phenylhydroxylamine, azoxybenzene, azobenzene and hydrazobenzene) are obtained.

Isopropyl group is nothing but propane with a hydrogen from middle C-atom removed. i.e., \(-CH(CH_{3})_{2}\). \(2\)-methylpentane has one isopropyl group.

\(2, 2, 3, 3\)-tetramethylpentane, \(2, 2\)-dimethylpentane and \(2, 2, 3\)-trimethylpentane do not contain an isopropyl group.

A functional isomer of Dimethyl ether is Ethanol.

It has the same chemical formula but different functional groups attached to them. e.g C₃H₆O

It has two functional isomers i.e.,

A deuterium nucleus consists of one proton and one neutron.

Benzene is not a saturated hydrocarbon.

Always suffix -ene contains double bond between carbon atoms and in case of -yne, it contains triple bond between carbon atoms, where as in case of -ane it contains single bond between carbon atoms. Saturated hydrocarons contain single bond between carbon atoms. Here benzene is not a saturated hydrocarbon.

In metallic conductors, the conductance is due to the flow of free mobile electrons and in electrolytic conductors, the conductance is due to the movement of ions in a solution of fused electrolyte.

\(\left[\mathrm{Fe}(\mathrm{CN})_6\right]^4 \rightarrow\)

No of unpaired electron \(=0\)

\(\left[\mathrm{MnCl}_4\right]^{2-} \rightarrow\)

No of unpaired electrons \(=5\)

\(\left[\mathrm{CoCl}_4\right]^{2-} \rightarrow\)

No of unpaired electrons \(=3\)

Note: The greater the number of unpaired electrons, greater the magnitude of magnetic moment. Hence the correct order will be \(\left[\mathrm{MnCl}_4\right]^{2-}>\left[\mathrm{CoCl}_4\right]^{2-}>\left[\mathrm{Fe}(\mathrm{CN})_6\right]^4\)

3,3-dimethylbutan-2-ol reacts with concentrated \(\mathrm{H}_2 \mathrm{SO}_4\) to form but-2,3-diene.

Hence, correct option is (b).

\(0.375 \mathrm{M}\) solution of \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)=0.375 \mathrm{moles}\)
of \(\left(\mathrm{CH}_{3} \text { COONa }\right)\) dissolved in
\(1000 \mathrm{ml}\) of solvent.
But according to question, we have to make a \(500 \mathrm{ml}\) solution of \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\)
\(\therefore\) Number of moles of sodium acetate in \(500 \mathrm{mL}\)
\(=\frac{0.375}{1000} \times 500\)
\(=0.1875 \mathrm{mole}\)
Molar mass of sodium acetate \(=82.0245 \mathrm{g} \mathrm{mole}^{-1}\) (Given)
\(\therefore\) Required mass of sodium acetate \(=\left(82.0245 \mathrm{g} \mathrm{mol}^{-1}\right)(0.1875 \mathrm{mole})\)
\(=15.38 \mathrm{g}\)