Chemistry Test - 21

When 1 litre of water is cooled from 4°C to 0°C, its volume increases.

• When water reaches 4°C the molecules have been pushed as close to one another as possible and the density of water becomes precisely 1.00 g/cm³
• When water freezes at 0°C due to the crystal structure the molecules arranged in some structured fashion so a little far apart ended up less dense - 0.93 g/cm3 - and so floats due to buoyancy.

According to Freundlich isotherm, at moderate pressure, extent of adsorption \(\left(\frac{x}{m}\right) \propto(p)^{\frac{1}{n}}\)

At moderate pressure,

\(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{k}(\mathrm{p})^{1 / \mathrm{n}} \)

\(\frac{\mathrm{x}}{\mathrm{m}} \propto(\mathrm{p})^{1 / \mathrm{n}} \ldots \text { (i) } \)

\(\frac{\mathrm{x}}{\mathrm{m}} \propto \mathrm{p}^{\mathrm{x}} \text { (Given in question) } \ldots \text { (ii) }\)

Compare Eqs. (i) and (ii),

\((p)^{1 / n} \propto p^x \)

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

There is a large difference in the boiling points of butanal and butan-1-ol due tointermolecular hydrogen bonding in butan-1-ol.

Butan-1-ol has polar O-H bond due to which it shows intermolecular H-bonding which is not possible in case of butanal due to absence of polar bond.

The simplest sugars which can not be further hydrolyzed are monosaccharides. Monosaccharides have a continuous carbon backbone; unlike disaccharides and polysaccharides they are not held together by oxygen-containing functional groups, which are more susceptible to hydrolysis.Monosaccharides are the building blocks of disaccharides (such as sucrose and lactose) and polysaccharides (such as cellulose and starch).

Graphite is one of the allotropes of carbon. Unlike diamond, it is an electrical conductor and a good lubricant. Graphite is a good conductor of electricity due to the presence of free valence electrons.

This property of graphite is due to the carbon atoms arranged in different layers and each atom is covalently bonded on three of its neighbouring atoms in the same layer. The fourth valence electrons of each atom are present between different layers and are free to move about. These free electrons in graphite make it a good conductor of electricity.

Alkyl halides develop colour on exposure to light.

• Alkyl bromides and iodides develop colour when exposed to light.
• Halogens have lower bond dissociation energy.
• This low enthalpy means that even at normal temperature or on exposure to sunlight, they decompose and form free ions, which are coloured and hence impart colour on exposure to light.

Given that: Mass of \({CO}({NO})_{3} .6 {H}_{2} {O}=30~g\)

Volume of solution \(=4.3 {~L}\)

We know that, Molarity \(=\frac{\text { Moles of Solute }}{\text { Volume of solution in litre }}\)

Molar mass of \({CO}({NO})_{3} \cdot 6 {H}_{2} {O}=59+2(14+3 \times 16)+6 \times 18=291 {~g}\) \({mol}^{-1}\)

Therefore, Moles of \({CO}({NO})_{3} \cdot 6 {H}_{2} {O}=\frac{\text{Mass}}{\text{Molar Mass}}\)

\(=\frac{30}{291} {~mol}\)\(=0.103 {~mol}\)

Therefore, Molarity \(=\frac{0.103 {~mol}}{4.3 L}\)\(=0.023 {M}\)

The fresh precipitate can be transformed in colloidal solution by peptization.

The dispersion of a freshly precipitated material into colloidal solution by the action of an electrolyte in solution is termed peptization. The electrolyte used is called a peptizing agent. For Example, sols obtained by peptization are: Freshly prepared ferric hydroxide on treatment with a small amount of ferric chloride solution at once forms a dark reddish brown solution. Ferric chloride acts as a peptizing agent.

As the temperature rises, more collisions start taking place which results in an increase in rate of the chemical reaction.

With rise in temperature, fraction of molecules possessing kinetic energy greater than the activation energy increases. Only such molecules are able to cause effective collisions and result in the formation of product. Thus, as the fraction of such molecules increases, rate of the reaction increases.

Assume the mass of benzene be \(30 {~g}\) in the total mass of the solution of \(100 {~g}\).

Mass of \({CCl}_{4}=(100-30) {g}\)

\(=70 {~g}\)

Molar mass of benzene \(\left({C}_{6} {H}_{6}\right)=(6 \times 12+6 \times 1) {g} {mol}^{-1}\)

\(=78 {~g} {~mol}^{-1}\)

Therefore,

Number of moles of \({C}_{6} {H}_{6} = \frac{\text{Mass}}{\text{Molar Mass}}\)

\(=\frac{30}{78} {~mol}\)

\(=0.3846 {~mol}\)

Molar mass of\(({CCl}_{4})=(1 \times 12+4 \times 355){~g} {~mol}^{-1}\)

\(=154 {~g} {~mol}^{-1}\)

Therefore,

Number of moles of\({CCl}_{4} = \frac{\text{Mass}}{\text{Molar Mass}}\)

\( =\frac{70}{154} ~mol\)

\(=0.4545 {~mol}\)

Therefore,

Mole fraction of benzene \(=\frac{\text { Number of moles of } C_{6} H_{6}}{\text { Number of moles of } C_{6} H_{6}+\text { Number of moles of } {CCl}_{4}}\)

\(=\frac{0.3846}{0.3846+0.4545}\)

\(=0.458\)