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Chemistry Test - 44
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  • Question 1
    1 / -0

    A saturated aqueous solution of sparingly soluble salt AB3 has the vapour pressure 0.08 mm lesser than the vapour pressure 17.33 mm of solvent at 25oC. The solubility product of AB3 is :

    Solution
    \(\mathrm{RLVP}=\frac{\Delta \mathrm{p}}{\mathrm{P}^{\circ}}=\frac{0.08}{17.33} \Rightarrow 4.616 \times 10^{-3}=\mathrm{i} \mathrm{X}_{\mathrm{B}}\)
    \(\mathrm{X}_{\mathrm{B}}=\) mole fraction of solute. \(\left(\mathrm{i}=4\right.\) for \(\mathrm{AB}_{3}\) salt \()\)
    \(\mathrm{X}_{\mathrm{B}}=\frac{4.616 \times 10^{-3}}{4}=1.154 \times 10^{-3}\)
    \(\mathrm{X}_{\mathrm{B}}=1.154 \times 10^{-3}=\frac{1}{1+\frac{\text { mole of } \mathrm{H}_{2} \mathrm{O}}{\text { moles of } \mathrm{salt}}}\)
    \(1+\frac{\text { mole of } \mathrm{H}_{2} \mathrm{O}}{\text { moles of salt }}=\frac{1}{1.154 \times 10^{-3}}=866.55\)
    \(\frac{\text { mole of } \mathrm{H}_{2} \mathrm{O}}{\text { moles of salt }}=865.55=\frac{55.55}{\text { mole of salt per litre }}\)
    \(M_{\text {salt }}=\frac{55.55}{865.55}=0.0641=S\)
    Salt \(\quad A B_{3} \rightleftharpoons \underset{S}{A}+3 \underset{3 S}{B}\)
    \(\mathrm{K}_{\mathrm{sp}}=27 \mathrm{S}^{4}=27(.0641)^{4}\)
    \(=4.56 \times 10^{-4}\)
    Hence, the correct option is (D)
  • Question 2
    1 / -0

    One desires to prepare a positively charged solution of silver iodide. This can be achieved by :

    Solution

    KI + AgNO3→ AgI + KNO3

    AgNO3 → Ag+ + NO3-

    AgI(s) + Ag+ → [AgI] +Ag+.

    Concepts :
    Main Concept :
    Multimolecular colloids, Macromolecular colloids and Associated colloids

    (a) Multimolecular colloids

    ● When on dissolution, atoms or smaller molecules of substances (having diameter less than 1nm) aggregate together to form particles of colloidal dimensions, the particles thus formed are called multimolecular colloids.

    ● In these sols the dispersed phase consists of aggregates of atoms or molecules with molecular size less than 1 nm.

    ● For example, sols of gold atoms and sulphur molecules. In these colloids, the particles are held together by Vander Waal's forces. They have usually lyophilic character.

    (b) Macromolecular colloids

    ● These are the substances having big size molecules (called macromolecules) which on dissolution form size in the colloidal range. Such substances are called macromolecular colloids.

    ● These macromolecules forming the dispersed phase are generally polymers having very high molecular masses.

    ● Naturally occurring macromolecules are starch, cellulose, proteins, enzymes, gelatin etc. Artificial macromolecules are synthetic polymers such as nylon, polythene, plastics, polystyrene etc.

    ● They have usually lyophobic character.

    (c) Associated colloids

    ● These are the substances which on dissolved in a medium behave as normal electrolytes at low concentration but behave, as colloidal particles at higher concentration due to the formation of aggregated particles. The aggregates particles thus formed are called micelles.

    ● Their molecules contain both lyophilic and lyophobic groups.

    Micelles

    ● Micelles are the cluster or aggregated particles formed by association of colloid in solution.

    ● The common examples of micelles are soaps and detergents.

    ● The formation of micelles takes place above a particular temperature called Kraft temperature (Tk) and above a particular concentration called critical micellization concentration (CMC).

    ● They are capable of forming ions.

    ● Micelles may contain as many as 100 molecules or more.

    ● For example sodium stearate (C17H35COONa) is a typical example of such type of molecules.

    Hence, the correct option is (B)

  • Question 3
    1 / -0

    What is the IUPAC name of the following compounds?

    Solution

    1-Chloro-3,3-dimethyl-1-phenyl butane (2obenzylic chloride)

    Concepts :
    Main Concept :
    IUPAC Rules for functional compoundsNaming Aldehydes
    The IUPAC system of nomenclature assigns a characteristic suffix -al to aldehydes. For example, H2C=O is methanal, more commonly called formaldehyde. Since an aldehyde carbonyl group must always lie at the end of a carbon chain, it is always is given the #1 location position in numbering and it is not necessary to include it in the name. There are several simple carbonyl containing compounds which have common names which are retained by IUPAC.

    Also, there is a common method for naming aldehydes and ketones. For aldehydes common parent chain names, similar to those used for carboxylic acids, are used and the suffix -aldehyde is added to the end. In common names of aldehydes, carbon atoms near the carbonyl group are often designated by Greek letters. The atom adjacent to the carbonyl function is alpha, the next removed is beta and so on.

    (-CHO) is attached to a ring the suffix -carbaldehyde is added to the name of the ring. The carbon attached to this moiety will get the #1 location number in naming the ring.
    Summary of Aldehyde Nomenclature rules

    I. Aldehydes take their name from their parent alkane chains. The -e is removed from the end and is replaced with -al.

    II. The aldehyde funtional group is given the #1 numbering location and this number is not included in the name.

    III. For the common name of aldehydes start with the common parent chain name and add the suffix -aldehyde. Substituent positions are shown with Greek letters.

    IV. When the -CHO functional group is attached to a ring the suffix -carbaldehyde is added, and the carbon attached to that group is C1.

    Naming Ketones

    The IUPAC system of nomenclature assigns a characteristic suffix of -one to ketones. A ketone carbonyl function may be located anywhere within a chain or ring, and its position is usually given by a location number. Chain numbering normally starts from the end nearest the carbonyl group. Very simple ketones, such as propanone and phenylethanone do not require a locator number, since there is only one possible site for a ketone carbonyl function

    The common names for ketones are formed by naming both alkyl groups attached to the carbonyl then adding the suffix -ketone. The attached alkyl groups are arranged in the name alphabetically.

    Summary of Ketone Nomenclature rules

    I. Ketones take their name from their parent alkane chains. The ending -e is removed and replaced with -one.

    II. The common name for ketones are simply the substituent groups listed alphabetically + ketone.

    III. Some common ketones are known by their generic names. Such as the fact that propanone is commonly referred to as acetone.Naming Aldehydes and Ketones in the Same MoleculeAs with many molecules with two or more functional groups, one is given priority while the other is named as a substituent. Because aldehydes have a higher priority than ketones, molecules which contain both functional groups are named as aldehydes and the ketone is named as an "oxo" substituent. It is not necessary to give the aldehyde functional group a location number, however, it is usually necessary to give a location number to the ketone.

    Naming Dialdehydes and Diketones

    For dialdehydes the location numbers for both carbonyls are omitted because the aldehyde functional groups are expected to occupy the ends of the parent chain. The ending -dial is added to the end of the parent chain name.For diketones both carbonyls require a location number. The ending -dione or -dial is added to the end of the parent chain.

    Naming Cyclic Ketones and Diketones

    In cyclic ketones the carbonyl group is assigned location position #1, and this number is not included in the name, unless more than one carbonyl group is present. The rest of the ring is numbered to give substituents the lowest possible location numbers. Remember the prefixcyclo is included before the parent chain name to indicate that it is in a ring. As with other ketones the -e ending is replaced with the -one to indicate the presence of a ketone.With cycloalkanes which contain two ketones both carbonyls need to be given a location numbers. Also, an -e is not removed from the end but the suffix -dione is added.

    Naming Carbonyls and Hydroxyls in the Same Molecule

    When and aldehyde or ketone is present in a molecule which also contains an alcohol functional group the carbonyl is given nomenclature priority by the IUPAC system. This means that the carbonyl is given the lowest possible location number and the appropriate nomenclature suffix is included. In the case of alcohols the OH is named as a hydroxyl substituent. However, the l in hydroxyl is generally removed

    Naming Carbonyls and Alkenes in the Same Molecule

    When and aldehyde or ketone is present in a molecule which also contains an alkene functional group the carbonyl is given nomenclature priority by the IUPAC system. This means that the carbonyl is given the lowest possible location number and the appropriate nomenclature suffix is included.When carbonyls are included with an alkene the following order is followed:(Location number of the alkene)-(Prefix name for the longest carbon chain minus the -ane ending)-(an -en ending to indicate the presence of an alkene)-(the location number of the carbonyl if a ketone is present)-(either an -one or and -anal ending). Remember that the carbonyl has priority so it should get the lowest possible location number. Also, remember that cis/tran or E/Z nomenclature for the alkene needs to be included if necessary.

    Functional groups and their priorities

    Table of Functional Group Priorities for Nomenclature


    Hence, the correct option is (D)

  • Question 4
    1 / -0

    1.44 g of Titanium (Ti) reacted with an excess of O2 and produced x gm of a non-stoichiometric compound Ti1.44O1. The value of x is :

    Solution
    \(\mathrm{Ti}+\mathrm{O}_{2} \longrightarrow \mathrm{T}_{1.44} \mathrm{O}\)

    (1.44 g)

    \((x) g\)

    Apply POAC to Ti

    \(\left(\frac{1.44 \mathrm{g}}{48}\right) \times 1=\left(\frac{\mathrm{xg}}{1.44 \times 48+16}\right) 1.44\)

    \(x=\frac{48 \times 1.44+16}{48}\)

    \(=1.44+\frac{16}{48}\)

    \(=1.44+0.33\)

    \(=1.77 \mathrm{g}\)

    Concepts :
    Main Concept :
    Principles of atom conservationThe principle implies Number of atoms remains constant. Lets understand this with an example.

    1g of a mixture of cuprous oxide and cupric oxide was quantitatively reduced to 0.839 g of metallic copper. What is the weight of cupric oxide in the original sample.

    The solution is:

    63.563 + 16 = 79.563

    127.126 + 16 = 143.126

    x*79.563 + y*143.126 = 1

    x*63.563 + y*127.126 = 0.839

    x = 0.006925497

    x*79.563 = 0.551 g cupric oxide
    Hence, the correct option is (C)
  • Question 5
    1 / -0

    Photoelectric emission is observed from a surface for frequencies \(\nu_{1}\) and \(\nu_{2}\) of the incident radiation \(\left(\nu_{1}>\nu_{2}\right)\). If the maximum kinetic energies of the photoelectrons in the two cases are in the ration \(1: \mathrm{k}\) then the threshold frequency \(\nu_{0}\) is given by :

    Solution

    Concepts :
    Main Concept :
    Examples on Photo electric effect

    Photoelectric effect
    The photoelectric experiment showed that the energy of the ejected electrons was proportional to the frequency of the illuminating light. This showed that whatever was knocking the electrons out had an energy proportional to light frequency
    The remarkable fact that the ejection energy was independent of the total energy of illumination showed that the interaction must be like that of a particle which gave all of its energy to the electron! This fit in well with Planck's hypothesis that light in the blackbody radiation experiment could exist only in discrete bundles with energy

    E = hν

     What is the maximum kinetic energy of the photo electrons emitted by a metal of work function v0 by a light of frequency v

    Step 1: Manipulate the equation E = hν − hν0 into the useful format.

    Step 2: Fill in the given data and the constants .

    Step 3: Calculate the maximum Energy of the electron emitted. It is the energy left after the minimum energy required to emit the electron

    Step 4: This information can be further used to find the stopping potential for photocurrent by eV = E where V is the stopping potential

    Example: 1. Electrons with a kinetic energy of 6.023 x 104 J/mol are evolved from the surface of a metal, when exposed to a radiation of wavelength of 600 nm (photoelectric effect). The minimum amount of energy required to remove an electron from the metal atom is :
    a) 2.3125 x 10-19 J b) 3 x 10-19 J c) 6.02 x 10-19 J d) 6.62 x 10-34 J

    When a radiation is passed on to the surface of a metal, some amount of energy is used in overcoming the attraction force on the electron and knocks it out from the atom. The remaining part is converted to kinetic energy.
    Therefore:
    Eradiation = Φ + KE
    Where:
    Φ = work function or ionization energy required to remove the electron from the atom.

    KE = kinetic energy
    Calculation of KE for one electron:

    In the question, KE for one mole of electrons is given. However, in the options, the energy values are given for one electron. Hence we have to find the KE value for one electron.

    From the data:

    KE for 6.023 x 1023 electrons (one mole) = 6.023 x 104 J

    KE for one electron = 6.023 x 104 / 6.023 x 1023 = 1 x 10-19 J

  • Question 6
    1 / -0

    Consider the three solvents of identical molar masses. Match their boiling point with their kb values

    \(\begin{array}{ccc}\text { Solvent } & \text { B.Pt. } & \mathrm{K}_{\mathrm{b}} \\ \mathrm{X} & 100^{\circ} \mathrm{C} & 0.92 \\ \mathrm{Y} & 27^{\circ} \mathrm{C} & 0.63 \\ \mathrm{Z} & 283^{\circ} \mathrm{C} & 0.53\end{array}\)

    Which of the following is the correct match between the boiling point and Kb values for X, Y, Z?

    Solution
    \(\mathrm{K}_{\mathrm{b}}=\frac{\mathrm{MRT}_{\mathrm{b}}^{\mathrm{s} 2}}{1000 \Delta \mathrm{H}_{\mathrm{vap}}}\)
    \(\mathrm{K}_{\mathrm{b}} \propto \frac{\mathrm{MT}_{\mathrm{b}}^{\circ}}{\left(\frac{\Delta \mathrm{H}_{\mathrm{vap}}}{\mathrm{T}_{\mathrm{b}}^{2}}\right)}\)
    \(\mathrm{K}_{\mathrm{b}} \propto \frac{\mathrm{T}_{\mathrm{b}}^{o}}{\Delta \mathrm{S}_{\mathrm{vap}}^{\circ}}\)
    \(\Delta \mathrm{S}^{\circ}\) will be same for three liquids changing to vapours for 1 mole
    \(\therefore \mathrm{K}_{\mathrm{b}} \propto \mathrm{T}_{\mathrm{b}}^{\circ}\)
    Hence, the correct option is (C)
  • Question 7
    1 / -0

    Acetamide is treated separately with the following reagents. Which of these would give methylamine ?

    Solution

  • Question 8
    1 / -0
    A solution of mercurous chloride on standing :
    Solution

    \(\mathrm{Hg}_{2} \mathrm{Cl}_{2} \quad\) standing \(\quad \mathrm{HgCl}_{2}+\mathrm{Hg} \downarrow\)

    (black colour)

    Hence option c is the correct answer.

  • Question 9
    1 / -0

    Which of the following cannot decompose on heating to give CO2 ?

    Solution

    Anhydrous sodium carbonate is stable to heat and does not decompose even when it is heated to redness. This is because sodium carbonate salt on heating with acids react to release carbon dioxide.

    Hence option B is the correct answer.

  • Question 10
    1 / -0

    Which of the following statement is correct?

    Solution

    Sodium carbonate, potassium carbonate and ammonium carbonate are soluble in water.

    Hence option C is the correct answer.

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