Question
Read the passage given below and answer the following questions: An ideal solution may be defined as the solution which obeys Raoult's law exactly over the entire range of concentration. The solutions for which vapour pressure is either higher or lower than that predicted by Raoult's law are called non-ideal solutions.Non-ideal solutions can show either positive or negative deviations from Raoult's law depending on whether the A-B interactions in solution are stronger or weaker than A - A and B - B interactions. The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which of the following solutions is/are ideal solution(s)?
  1. Bromoethane and iodoethane.
  2. Acetone and chloroform
  3. Benzene and acetone
  4. n-heptane and n-hexane
  1. Only I
  2. I and II
  3. II and III
  4. I and Iv
  1. For which of the following solutions $\Delta\text{H}_{\text{mix}}$ and $\Delta\text{V}_{\text{mix}}$ is negative?
  1. Acetone and aniline
  2. Ethyl alcohol and cyclohexane
  3. Acetone and CS2
  4. Benzene and toluene
  1. Which of the following is not true for positive deviations?
  1. The A-B interactions in solution are weaker than the A - A and B - B interactions.
  2. $\text{P}_\text{A}<\text{P}\mathring{\text{A}}\ \text{x}_\text{A}$ and $\text{P}_\text{B}<\text{P}\mathring{\text{B}}\ \text{x}_\text{B}$
  3. Carbon tetrachloride and chloroform mixture is an example of positive deviations.
  4. All of these.
  1. For water and nitric acid mixture, which of the given graph is correct?
  1.  
  1.  
  1. Both of these
  2. None of these
  1. Water-HCI mixture.
  1. Shows positive deviations.
  2. Forms minimum boiling azeotrope.
  3. Shows negative deviations.
  4. Forms maximum boiling azeotrope.
  1. I and II
  2. I and III
  3. I and IV
  4. III and IV

Answer

  1. (d) I and Iv
Explanation:

II represents negative deviations and III represents positive deviations.
  1. (a) Acetone and aniline
Explanation:

Acetone and aniline mixture represents negative deviations from Raoult's law, hence for this mixture,

$\Delta\text{H}_{\text{mix}}$ and $\Delta\text{V}_{\text{mix}}$ is negative.
  1. (b) $\text{P}_\text{A}<\text{P}\mathring{\text{A}}\ \text{x}_\text{A}$ and $\text{P}_\text{B}<\text{P}\mathring{\text{B}}\ \text{x}_\text{B}$
Explanation:

For positive deviations $\text{P}_\text{A}<\text{P}\mathring{\text{A}}\ \text{x}_\text{A}$ and $\text{P}_\text{B}<\text{P}\mathring{\text{B}}\ \text{x}_\text{B}$
  1. (b)
Explanation:

Water and nitric acid mixture shows negative deviations from Raoult's law, hence

$\text{P}_\text{A}<\text{P}\mathring{\text{A}}\ \text{x}_\text{A}$ and $\text{P}_\text{B}<\text{P}\mathring{\text{B}}\ \text{x}_\text{B}$
  1. (d) III and IV
Explanation:

Water-HCI mixture shows negative deviations from Raoult's law and solutions showing negative deviations from ideal behaviour form maximum boiling azeotrope.

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Nemst equation relates the reduction potential of an electrochemical reaction to the standard potential and activities of the chemical species undergoing oxidation and reduction.
Let us consider the reaction, $\text{M}^{\text{n+}}_{(\text{aq})}\xrightarrow{\ \ \ \ \ \ \ \ }\text{nM}_\text{(s)}$
For this reaction, the electrode potential measured with respect to standard hydrogen electrode can be given as
$\text{E}_{\Big(\frac{\text{M}^{\text{n+}}}{\text{M}}\Big)}=\text{E}^\circ_{\Big(\frac{\text{M}^\text{n+}}{\text{M}}\Big)}-\frac{\text{RT}}{\text{nF}}\text{ln}\frac{[\text{M}]}{[\text{M}^{\text{n}+}]}$
In these questions (Q. No. i-iv), a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.
  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: For concentration cell, $\text{Zn}_{(\text{s})}|\text{ Zn}^{2+}_{\text{(aq)}}||\text{ Zn}^{2+}_{(\text{aq})}|\text{ Zn}\\\ \ \ \ \ \ \ \ \ \ \ \ \text{C}_1\ \ \ \ \ \ \ \ \text{C}_2$
For spontaneous cell reaction, $C_1 < C_2$
Reason: For concentration cell, $\text{E}_\text{cell}=\frac{\text{RT}}{\text{nF}}\log\frac{\text{C}_2}{\text{C}_1}$
For spontaneous reaction, $\text{E}_\text{cell}=+\text{ve}\Rightarrow\text{C}_2>\text{C}_1$
  1. Assertion: For the cell reaction, $\text{Zn}_{(\text{s})}+\text{Cu}^{2+}_{(\text{aq})}\xrightarrow{\ \ \ \ \ }\text{Zn}^{2+}_{(\text{aq})}+\text{Cu}_{(\text{s})}$ voltmeter gives zero reading at equilibrium.
Reason: At the equilibrium, there is no change in concentration of Cu2+ and Zn2+ ions.
  1. Assertion: The Nernst equation gives the concentration dependence of emf of the cell.
Reason: In a cell, current flows from cathode to anode.
  1. Assertion: Increase in the concentration of copper half cell in a cell, increases the emfofthe cell.
Reason: $\text{E}_\text{cell}=\text{E}^\circ_\text{cell}+\frac{0.059}{2}\log\frac{[\text{Cu}^{2+}]}{[\text{Zn}^{2+}]}$
  1. Assertion: Electrode potential for the electrode $\frac{\text{Mn}^+}{\text{Mn}}$ with concentration is given by the expression under STP conditions.
$\text{E}=\text{E}^\circ+\frac{0.059}{\text{n}}\log[\text{Mn}^{+}]$
Reason: STP conditions require the temperature to be 273K.
Read the passage given below and answer the following questions: The order of reactivity towards $S_N1$ reaction depends upon the stability of carbocation in the first step. Greater the stability of the carbocation, greater will be its ease of formation from alkyl halide and hence faster will be the rate of the reaction. As we know, $3^\circ$ carbocation is most stable, therefore, the tert-alkyl that halides will undergo $S_N1$ reaction very fast. For example, it has been observed that the reaction $(CH_3)_3CBr$ with $OH^-$ ion to give 2-methyl-2-propanol is about I million times as fast as the corresponding reaction of the methyl bromide to give methanol. The primary alkyl halides always react predominantly by $S_N2$ mechanism. On the other hand, the tertiary alkyl halides react predominantly by $S_N1$ mechanism. Secondary alkyl halides may react by either mechanism or by both the mechanisms without much preference depending upon the nature of the nucleophile and solvent. In these questions (Q. No. i-tv), a statement of assertion followed by a statement of reason is given. Choose tile correct answer out of tile following choices.
  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: Low concentration of nudeophile favours $S_N1$ mechanism.
Reason: $2^\circ$ alkyl halides are less reactive than $1^\circ$ towards $S_N1$ reactions.
  1. Assertion: Polar solvent slows down $S_N2$ reactions.
Reason: $CH_3-Br$ is less reactive than $CH_3Cl.$
  1. Assertion: Benzyl bromide when kept in acetone- water it produces benzyl alcohol.
Reason: The reaction follows $S_N2$ mechanism.
  1. Assertion: Rate of hydrolysis of methyl chloride to methanol is higher in DMF than in water.
Reason: Hydrolysis of methyl chloride follows second order kinetics.
  1. Assertion: $S_N1$ reaction is carried out in the presence of a polar protic solvent.
Reason: A polar protic solvent increases the stability of carbocation due to solvation.
Read the passage given below and answer the following questions:
Valence bond theory considers the bonding between the metal ion and the ligands as purely covalent. On the other hand, crystal field theory considers the metal-ligand bond to be ionic arising from electrostatic interaction between the metal ion and the ligands. In coordination compounds, the interaction between the ligand and the metal ion causes the five d-orbitals to split-up. This is called crystal field splitting and the energy difference between the two sets of energy level is called crystal field splitting energy. The crystal field splitting energy $(\Delta_0)$ depends upon the nature of the ligand. The actual configuration of complexes is divided by the relative values of $\Delta_0$ and P (pairing energy).
If $\Delta_0<\text{P},$ then complex will be high spin.
If $\Delta_0>\text{P},$ then complex will be low spin
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which of the following ligand has lowest $\Delta_0$ value?
  1. $CN^-$
  2. $CO$
  3. $F^-$
  4. $NH_3$
  1. The crystal field splitting energy for octahedral $(\Delta_0)$ and tetrahedral $(\Delta_t)$ complex is related as:
  1. $\Delta_\text{t}=\frac{1}{2}\Delta_0$
  2. $\Delta_\text{t}=\frac{4}{9}\Delta_0$
  3. $\Delta_\text{t}=\frac{3}{5}\Delta_0$
  4. $\Delta_\text{t}=\frac{2}{5}\Delta_0$
  1. On the basis of crystal field theory, the electronic configuration of $d_4$ in two situations : (i) t.0 > P and (ii) t.0
  (i) (ii)
(a) $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$ $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$
(b) $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$ $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$
(c) $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$ $\text{t}^3_{2\text{g}}\text{e}^1_\text{g}$
(d) $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$ $\text{t}^4_{2\text{g}}\text{e}^0_\text{g}$
  1. Using crystal field theory, calculate magnetic moment of central metal ion of $[FeF_6]^{4-}$.
  1. 1.79B.M.
  2. 2.83B.M.
  3. 3.85B.M.
  4. 4.9B.M.
  1. Electronic configuration of d-orbitals in $[Ti(H_2O)_6]^{3+}$​​​​​​​ ion in an octahedral crystal field is:
  1. $\text{t}^1_{2\text{g}}\text{e}^0_\text{g}$
  2. $\text{t}^2_{2\text{g}}\text{e}^0_\text{g}$
  3. $\text{t}^0_{2\text{g}}\text{e}^1_\text{g}$
  4. $\text{t}^1_{2\text{g}}\text{e}^1_\text{g}$
Read the passage given below and answer the following questions: Aniline activates the benzene ring by increasing electron density at ortho- and para-positions. Hence, it is o-, p-directing. -NH2 group strongly activates the ring therefore it is difficult to stop the reaction at monosubstitution stage. Among electrophilic substitution reaction, direct nitration of aniline is not done to get o- and p-nitroaniline because lone pair of electrons present at nitrogen atom will accept proton from nitrating mixture to give anilinium ion which is meta-directing. Aniline with $NaNO_2$ and HCI forms benzene diazonium chloride at very low temperature. Aromatic amines react with nitrous acid to form a yellow oily liquid known as N-nitrosoamines. A statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: Nitrating mixture used for carrying out nitration of benzene consists of cone. $HNO_3$ + cone. $H_2SO_4$.
Reason: In presence of $H_2SO_4, HNO_3$ acts as a base and produces $\text{NO}^+_2$ ions.
  1. Assertion: Anilinium chloride is more acidic than ammonium chloride.
Reason: Anilinium ion is not resonance-stabilised.
  1. Assertion: Nitrobenzene can be prepared from benzene by using mixture of cone. $HNO_3$ and cone. $H_2SO_4$.
Reason: In the mixture, $H_2SO_4$ act as a acid.
  1. Assertion: In strongly acidic solution, aniline becomes less reactive towards electrophilic reagents.
Reason: The amino group being completely protonated in strongly acidic solution, the lone pair of electrons on the nitrogen is no longer available for resonance.
  1. Assertion: Nitration of aniline can be done conveniently by protecting $-NH_2$ group through acetylation.
Reason: Acetylation of aniline results in the increase of electron density in the benzene ring.
In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Lyophilic sols are prepared by directly mixing the substance with the dispersion medium.
Reason: Lyophobic sols can not be prepared by simply mixing the substance with the dispersion medium.
Read the passage given below and answer the following questions:
At the freezing point of a solvent, the solid and the liquid are in equilibrium. Therefore, a solution will freeze when its vapour pressure becomes equal to the vapour pressure of the pure solid solvent. It has been observed that when a non-volatile solute is added to a solvent, the freezing point of the solution is always lower than that of the pure solvent. Depression in freezing point can be given as, $\Delta\text{T}_\text{f}=\text{K}_\text{f}\text{m}$ Where, $K_f =$ Molal freezing point depression constant or we can write, $\Delta\text{T}_\text{f}=\frac{\text{K}_\text{f}\times\text{W}_\text{B}\times1000}{\text{W}_A\times\text{M}_\text{B}}$
a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
  1. Assertion: $0.1M$ solution of glucose has same depression in the freezing point as $0.1M$ solution of urea.
Reason: $K_f $ for both has same value.
  1. Assertion: Increasing pressure on pure water decreases its freezing point.
Reason: Density of water is maximum at $273K.$
  1. Assertion: Larger the value of cryoscopic constant of the solvent, lesser will be the freezing point of the solution.
Reason: Extent of depression in the freezing point depends on the nature of the solvent.
  1. Assertion: The water pouch of instant cold pack for treating athletic injuries breaks when squeezed and NH4N03 dissolves thus lowering the temperature.
Reason: Addition of non-volatile solute into solvent results into depression of freezing point of solvent.
  1. Assertion: If a non-volatile solute is mixed in a solution then elevation in boiling point and depression in freezing point both wiII be same.
Reason: Elevation in boiling point and depression in freezing point both depend on number of particles of solute.
Write detailed note on: Starch
Read the passage given below and answer the following questions:
Electron microscopic study of crystal defects enables us not only to reveal various structural imperfections, but also to discover their formation, mechanisms and to understand their effects on the properties of solid materials. There are commonly two types of imperfections: electronic imperfections and atomic imperfections or point defects.Electronic imperfections correspond to defects in ionic crystal due to the electrons. Atomic imperfections or point defects correspond to the irregularity of atoms around a point or an atom. The point defects in ionic crystals may be classified as: defects in stoichiometric crystals, defects in non-stoichiometric crystals and impurity defects. In stoichiometric crystals, generally two types of defects are observed: Schottky defect and Frankel defect. Schottky defect arises when some of the atoms or ions are missing from their normal lattice sites. Due to the schottky defect, density of ionic crystals decreases markedly. For example NaCl, KCl, CsCl, AgBr ionic solids have schottky defects. It has been observed that in NaCl, there are about $10^6$ Schottky pairs per cm 'at room temperature. Frankel defect arises when an ion is missing from its normal position and occupies an interstitial site between the lattice points. It does not affect the density of the crystals. In non-stoichiometric crystals, two types of defects are there ; metal excess defects and metal deficient defects. In metal excess defect, the positive ions are in excess whereas in metal deficient defects, number of positive ions are less than the negative ions. Impurity defects arise due to presence of some impurity ions at the lattice sites.
In these questions (Q. No. i-iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
  1. Assertion and reason both are correct statements, and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements, but reason is not correct explanation for assertion.
  3. Assertion is correct statement, but reason is wrong statement.
  4. Assertion is wrong statement, but reason is correct statement.
  1. Assertion: In any ionic solid (MX) with Schottky defects, the number of positive and negative ions are same.
Reason: Equal number of cation and anion vacancies are present.
  1. Assertion: Due to Frenkel defect, there is no effect on the density of the crystalline solid.
Reason: In Frenkel defect, no cation or anion leaves the crystal.
  1. Assertion: The presence of a large number of Schottky defects in NaCl lowers its density.
Reason: In NaCl, there are approximately $10^6$ Schottky pairs per $cm^3$ at room temperature.
  1. Assertion: No compound has both Schottky and Frenkel defects.
Reason: Schottky defects change the density of the solid.
  1. Assertion: NaCl and KCl show metal excess defect.
Reason: Zinc oxide is white in colour at room temperature and on heating it loses oxygen and turns yellow due to metal excess defect.
Read the passage given below and answer the following questions:
A primary alkyl halide (A) $C_4H_9Br$ reacted with akoholic KOH to give compound (B). Compound (B) is reacted with HBr to give compound (C) which is an isomer of (A). When (A) reacted with sodium metal, it gave a compound (D) $C_8H_{18}$ that is different than the compound obtained when n-butyl bromide reacted with sodium metal.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Compound (A) is:
  1. $CH_3CH_2CH_2CH_2Br$
  2. $\text{CH}_3\text{CH}-\text{CH}_2\text{Br}\\\ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \text{CH}_3$
  3. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{Br} \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
  4. $CH_3CH_2CH_2Br$
  1. Which type of isomerism is present in compound (A) and (C)?
  1. Positional
  2. Functional
  3. Chain
  4. Both (a) and (c)
  1. Identify compound (B).
  1. $\text{CH}_3-\text{C}=\text{CH}_2 \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \mid \\ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
  2. $CH_3– CH = CH – CH_3$
  3. $CH_3– CH_2 – CH = CH_2$
  4. None of these.
  1. IUPAC name of compound (D) is:
  1. N - octane
  2. 2, 5 - dimethylhexane
  3. 2 - methylheptane
  4. 3, 4 - dimethyl hexane.
  1. When compoound (C) is treated with ale. KOH and then treated with presence of peroxide, the compound obtained is:
  1. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{C}-\text{Br} \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
  2. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}-\text{CH}_2-\text{Br}$
  3. $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$
  4. $$$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}-\text{Br}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
Read the passage given below and answer the following questions:
Arrangement of ligands in order of their ability to cause splitting $(\Delta)$ is called spectrochemical series. Ligands which cause large splitting (large $\Delta$) are called strong field ligands while those which cause small splitting (small $\Delta$) are called weak field ligands. When strong field ligands approach metal atom/ ion, the value of $\Delta_0$ is large, so that electrons are forced to get paired up in lower energy $t_{2g}$ orbitals. Hence, a low-spin complex is resulted from strong field ligand. When weak field ligands approach metal atom/ ion, the value of $\Delta_0$ is small, so that electrons enter high energy $e_g$ orbitals rather than pairing in low energy $t_{2g}$ orbitals. Hence, a high-spin complex is resulted from weak field ligands. Strong field ligands have tendency to form inner orbital complexes by forcing the electrons to pair up. Whereas weak field ligands have tendency to form outer orbital complex because inner electrons generally do not pair up.
In these questions (Q. No. i-iv), a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.
  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4. Assertion is wrong statement but reason is correct statement.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Assertion: In tetrahedral coordination entity formation, the d orbital splitting is inverted and is smaller as compared to the octahedral field splitting.
Reason: Spectrochemical series is based on the absorption of light by complexes with different ligands.
  1. Assertion: In high spin situation, configuration of $d^5$ ions will be $\text{t}^3_{2\text{g}}\text{e}^2_\text{g}.$
Reason: In high spin situation, pairing energy is less than crystal field energy.
  1. Assertion: $F^-$ ion is a weak field ligand and fonns outer orbital complex.
Reason: $F^-$ ion cannot force the electrons of $d_{z^2}$ and $d_{x^2-y^2}$ orbitals of the inner shell to occupy $d_{xy}, d_{yz}$ and $d_{zx}$ orbitals of the same shell.
  1. Assertion: The crystal field model is successful in explaining the formation, structures, colour and magnetic properties of coordination compounds.
Reason: In spectrochemical series, ligands are arranged in a series of increasing field strength.
  1. Assertion: $NF_3$ is a weaker ligand than $N(CH_3)_3$.
Reason: $NF_3$ ionizes to give $F^-$ ions in aqueous solution.