Chemistry Test-14

Raoult’s law suggests that the difference between the vapour pressure of the pure solvent and that of solution increases as the mole fraction of the solvent decreases.

Mole fraction plays a role in this where it is defined as the number of moles of that component in that component in that particular phase.

Now, according to the data given vapour pressure of liquid \(A\) is \(400 mm\) and vapour pressure of liquid \(B\) is \(600 mm\).

That is \(P_A^0=400 mm\) and \(P_B{ }^0=600 mm\)

Thus, total vapour pressure of the ideal solution according to Raoult's law is as follows,

\(P=\chi_A P_A^0+\chi_B P_B^0 \dots......(1)\)

where, \(\chi_A\) and \(\chi_B\) are mole fraction of solution \(A\) and \(B\) respectively.

Now, since there is \(1\) mole of \(A\) and let \(x\) be the number of moles \(B\) then total moles will be \(1+x\)

Now substituting all these values in equation \((1)\), we get

\(550=\frac{1}{1+x}(400)+\frac{x}{1+x}(600)\)

Now rearranging this above equation we get,

\(550(1+x)=400+600 x\)

\(600 x-550 x=550-400\)

\(\Rightarrow 50 x=150\)

\(\Rightarrow x=3\)

In decreasing order of basic strength in the gas phase,

In the gas phase, there is no hydrogen bonding. Therefore stabilization due to hydrogen bonding is not there. Thus, the only effect to determine the strength is the inductive effect. The +I effect increases with an increase in the alkyl group. Therefore the basic strength will be the highest in (C₂H₅)₃N and least in NH₃.Therefore the increasing order of basic strength in the gas phase will be,

The magnitude of the electrode potential of a metal is a measure of its relative tendency to lose or gain electrons. i.e., it is a measure of the relative tendency to undergo oxidation (loss of electrons) or reduction (gain of electrons).

M → Mⁿ⁺ + ne^– (oxidation potential)

Mⁿ⁺ + ne^– → M (reduction potential)

According to the question,

(i)\(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ; \Delta_{\mathrm{c}} \mathrm{H}^{\Theta}=-890.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

(ii) \(\mathrm{C}\) (s) \(+2 \mathrm{O}_{2}\) (g) \(\rightarrow \mathrm{CO}_{2}(\mathrm{~g}) ; \Delta_{\mathrm{c}} \mathrm{H}^{\Theta}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

(iii) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}\) (g) \(\rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ; \Delta_{\mathrm{c}} \mathrm{H}^{\Theta}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Thus, the desired equation is the one that represents the formation of \(\mathrm{CH}_{4}(\mathrm{~g})\) that is as follows:

\(\mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{4}(\mathrm{~g}) ; \Delta_{\mathrm{f}} \mathrm{H}_{\mathrm{CH}_{4}}=\Delta_{\mathrm{c}} \mathrm{H}_{\mathrm{c}}+2 \Delta_{\mathrm{c}} \mathrm{H}_{\mathrm{H}_{2}}-\Delta_{\mathrm{c}} \mathrm{H}_{\mathrm{CO}_{2}}\)

Substituting the values in the above formula :

Enthalpy of formation \(\mathrm{CH}_{4}(\mathrm{g})=(-393.5)+2 \times(-285.8)-(-890.3)=-74.8 \mathrm{kJmol}^{-1}\)

Given: Current, \(I=1 \mu A =10^{-6} A\)

We know that \(Q=n e, t=1 s\)

Where \(n\) is the number of electrons and e is the charge on 1 electron and \(e=1.6 \times 10^{-19} C\)

\(n=\frac{I}{e}=\frac{10^{-6}}{1.6 \times 10^{-19}}=6.25 \times 10^{12}\) Electrons

A mixture of 1°, 2° and 3° amines can be separated by Hinsberg’s method.

Mixture of primary, secondary and tertiary amines can be separated by Hinsberg's method and fractional distillation.
They give different products with Hinsberg's. These products differ in solubility. Hence, they can be separated with Hinsberg's method. They differ in boiling points. Hence they can be separated by fractional distillation.

​Every orbital is has a unique and characteristic shape.

s-orbital is spherically shaped.

p-orbital is of dumbbell shape.

d-orbital is of double dumbbell shape.

f-orbital is of eight lobed dumbbell shape.

Given equation:

\(2 HI \rightleftharpoons H _{2}( g )+ I _{2}( g )\)

constant \(K\) at \(25^{\circ} C\) \(\Delta G =1.7 kJ\)

we know \(\Delta G = -RT ln K\)

\(1.7 \times 10^{3}=-2.303 \log K ( RT )\)

\(1700=-2.303(8.314)(298) \log K\)

\(\log K =-\frac{1700}{2.303 \times 8.314 \times 298}=\frac{1700}{5706}\)

So \(\log K =-0.29\)

\(K =0.5 \)

Sequence of energy levels of molecular orbitals is not correct for the remaining molecules \(L i_{2}, B e_{2}, B_{2}, C_{2}, N_{2}\)

For instance, it has been observed experimentally that for molecules such as \(B_{2}, C_{2}, N_{2}\) etc.

The increasing order of energies of various molecular orbitals is

\(\sigma 1 s<\sigma^{\star} 1 s<\sigma 2 s<\sigma^{*} 2 s<\left(\pi 2 p_{x}=\pi 2 p_{y}\right)<\sigma 2 p_{z}<\left(\pi^{\star} 2 p_{x}=\pi^{*} 2 p_{y}\right)<\sigma^{\star} 2 p_{z}\)

The important characteristic feature of this order is that the energy of \(\sigma 2 {pz}\) molecular orbital is higher than that of \(\pi 2 p_{x}\) and \(\pi 2 p_{y}\) molecular orbitals.

A basic event in protein synthesis is the creation of a/an mRNA (Messenger Ribonucleic Acid) copy.The mRNA provides the template for the Genetic Code.A gene is used to build a protein using the following two-step process i.e. Transcription and Translation.

• Transcription:It is the process by which DNA is copied to mRNA, which carries the information needed for protein synthesis.
• Translation:It is the process that takes the information passed from DNA as messenger RNA and turns it into a series of amino acids.