Chemistry Test - 53

Acetylene on reaction with \({NaNH}_{2}\) forms sodium salt.

The sp hybridized carbon atom in acetylene has higher

electronegativity as compared to \(s p^{2}\) hybridized carbon atom in ethylene and \(s p^{3}\) hybridized carbon atom in ethane. The carbon atom of \(C-H\) bond in acetylene attracts the shared electron pair towards itself to a greater extent and the hydrogen atom can be easily liberated as the proton. Hence, the hydrogen atom attached to the triply bonded carbon atom in alkynes has appreciable acidic character.

When acetylene reacts with \({NaNH}_{2}\) which is the strong base reacts to form a sodium salt.

When the pure solvent diffuses out of the solution through the semi-permeable membrane then the process is called reverse osmosis. High pressure is applied, forcing water from a concentrated aqueous solution like seawater to pure water side through the semi-permeable membrane.

Baeyer's reagent (alk. \(\mathrm{KMnO}_{4}\) ) which is pink in colour decolourises due to the presence of unsaturation.

Thus, it shows the presence of unsaturation in an organic compound.

DNA has a double-stranded structure that is helical in orientation.In a DNA strand, the nucleotides are arranged in a manner that adenine is always paired with thymine and cytosine is always paired with guanine. Adenine and thymine are joined by two hydrogen bonds and guanine and cytosine are always joined by three hydrogen bonds.The Chargaff’s rule of base pairs states that the base pairs are present in a one to one ratio in a DNA segment. According to this, if there are \(120\) adenine, there have to be \(120\) thymines also present. If there are \(120\) cytosines, there should be \(120\) guanines also present. So there will be \(480\) nucleotides present in the given segment.

Given that:

Weight of solute \(=1.00~g\)

Molar Mass of the solute \(=250~g~mol^{-1}\)

Weight of solvent \(=51.2~g=0.0512~kg\)

We know that:

Molality of non- electrolyte solute \( =\frac{\frac{\text { weight of solute in gram }}{\text { molecular weight of solute }}}{\text { weight of solvent in } {kg}}\)

\(=\frac{\frac{1}{250}}{0.0512}\)

\(=\frac{1}{250 \times 0.0512}\)

\(=0.0781 {~m}\)

As we know:

\(\Delta {T}_{{f}}={K}_{{f}} \times\) molality of solution

\(=5.12 \times 0.0781\)

\(=0.4 {~K}\)

Physical and chemical adsorption resemble only in effect of surface area, as:

(a) Force of attraction is more in case of chemical adsorption.

(b) Enthalpy of adsorption is more negative for chemical adsorption.

(c) Chemical adsorption depends on activation energy. Thus, it initially increases then decreases for increase in temperature. While physical adsorption only decreases with the increase in temperature.

(d) Physical adsorption is multilayered while chemical adsorption is single layer adsorption.

For first order reaction, 

\([A]=[A_0] e ^{-kt}\)

Where, k = First order rate constant

[A]₀ = Initial concentration

[A] = Concentration at time 't'

From the above equation we can draw:

Thus we can say that the concentration of reactants will exponentially decrease with time.

Given, The solution contains \(10 \mathrm{~g}\) per \(d m^{3}\) urea.

And, we know that: The chemical formula of urea is \(\mathrm{CH}_{4} \mathrm{~N}_{2} \mathrm{O}\).

So, molecular mass of urea \(=12+4 \times(1)+2 \times(14)+16\)\(=60 \mathrm{~gmol}^{-1}\)

So, the molarity of urea = \(\frac{\text{mass concentration}}{\text{molar mass}}\) \(=\left(\frac{10 g / d m^{3}}{60}\right)\)

\(=\frac{1}{6} m o l / d m^{3}\) 

Let's assume the molar mass of the non-volatile solution is '\(\mathrm{m}\)'.

So, the molarity of non- volatile solute =\(\frac{\text{mass concentration}}{\text{molar mass}}\) \(=\frac{50 g / d m^{3}}{m}\)

Both solutions are isotonic with each other means concentration of both solutions are same:

Molarity of urea = Molarity of non-volatile solution

\(\frac{1}{6} \mathrm{~mol} / \mathrm{dm}^{3}=\frac{50 \mathrm{~g} / \mathrm{dm}^{3}}{\mathrm{~m}}\)

By solving the above equation we get the value of '\(m\)' as:

\(m=50 \times 6 \mathrm{~gmol}^{-1}\)

\(=300 \mathrm{~gmol}^{-1}\)