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Solutions — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistrySolutions4 Marks
Question
Read the passage given below and answer the following questions: Boiling point or freezing point of liquid solution would be affected by the dissolved solids in the liquid phase. A soluble solid in solution has the effect of raising its boiling point and depressing its freezing point. The addition of non-volatile substances to a solvent decreases the vapor pressure and the added solute particles affect the formation of pure solvent crystals. According to many researches the decrease in freezing point directly correlated to the concentration of solutes dissolved in the solvent. This phenomenon is expressed as freezing point depression and it is useful for several applications such as freeze concentration of liquid food and to find the molar mass of an unknown solute in the solution. Freeze concentration is a high quality liquid food concentration method where water is removed by forming ice crystals. This is done by cooling the liquid food below the freezing point of the solution. The freezing point depression is referred as a colligative property and it is proportional to the molar concentration of the solution (m), along with vapor pressure lowering, boiling point elevation, and osmotic pressure. These are physical characteristics of solutions that depend only on the identity of the solvent and the concentration of the solute. The characters are not depending on the solute’s identity.
  1. When a non volatile solid is added to pure water it will:
  1. boil above 100°C and freeze above 0°C
  2. boil below 100°C and freeze above 0°C
  3. boil above 100°C and freeze below 0°C
  4. boil below 100°C and freeze below 0°C
  1. Colligative properties are:
  1. dependent only on the concentration of the solute and independent of the solvent’s and solute’s identity.
  2. dependent only on the identity of the solute and the concentration of the solute and independent of the solvent's identity.
  3. dependent on the identity of the solvent and solute and thus on the concentration of the solute.
  4. dependent only on the identity of the solvent and the concentration of the solute and independent of the solute’s identity.
  1. Assume three samples of juices A, B and C have glucose as the only sugar present in them. The concentration of sample A, B and C are 0.1M, 5M and 0.2 M respectively. Freezing point will be highest for the fruit juice:
  1. A
  2. B
  3. C
  4. All have same freezing point
  1. Identify which of the following is a colligative property:
  1. freezing point
  2. boiling point
  3. osmotic pressure
  4. all of the above
✓

Answer

  1. (b) boil below 100°C and freeze above 0°C
  2. (d) dependent only on the identity of the solvent and the concentration of the solute and independent of the solute’s identity.
  3. (a) A
  4. (c) osmotic pressure

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Read the passage given below and answer the following questions:

Transition metal oxides are compounds fanned by the reaction of metals with oxygen at high temperature. The highest oxidation number in the oxides coincides with the group number. In vanadium, there is a gradual change from the basic V2O3 to less basic V2O4 and to amphoteric V2O5· V2O4 dissolves in acids to give VO2+ salts. Transition metal oxides are commonly utilized for their catalytic activity and semi conductive properties. Transition metal oxides are also frequently used as pigments in paints and plastic. Most notably titatnium dioxide. One of the earliest application of transition metal oxides to chemical industry involved the use of vanadium oxide for catalytic oxidation of sulfur dioxide to sulphuric acid. Since then, many other applications have emerged, which include benzene oxidation to maleic anhydride on vandium oxides; cyclohexane oxidation to adipic acid on cobalt oxides. An important property of the catalyst material used in these processes is the ability of transition metals to change their oxidation state under a given chemical potential of reductants and oxidants.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. Which oxide of vanadium is most likely to be basic and ionic?
  1. VO
  2. V2O3
  3. VO2
  4. V2O5
  1. Vanadyl ion is:
  1. $\text{VO}^{2+}$
  2. $\text{VO}^{+}_2$
  3. $\text{V}_{2}\text{O}^+$
  4. $\text{VO}^{3-}_4$
  1. Which of the following statements is false?
  1. With fluorine vanadium can form VF5.
  2. With chlorine vanadium can form VCl5.
  3. Vanadium exhibits highest oxidation state in oxohalides VOCl3, VOBr3 and fluoride VF5.
  4. With iodine vanadium cannot form Vl5 due to oxidising power of V5+ and reducing nature of I-.
  1. The oxidation state of vanadium in V2O5 is:
  1. $\frac{+5}{2}$

  2. +7
  3. +5
  4. +6
  1. Identify the oxidising agent in the following reaction.

V2O5 + 5Ca → 2V + 5CaO

  1. V2O5
  2. Ca
  3. V
  4. None of these.
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Read the passage given below and answer the following questions:
The amines are basic in nature due to the presence of a lone pair of electron on N-atom of the -NH2 group, which it can donate to electron deficient compounds. Aliphatic amines are stronger bases than NH3 because of the +I effect of the alkyl groups. Greater the number of alkyl groups attached to N-atom, higher is the electron density on it and more will be the basicity. Thus, the order of basic nature of amines is expected to be 3º > 2º > 1º, however the observed order is 2º > 1º > 3º. This is explained on the basis of crowding on N-atom of the amine by alkyl groups which hinders the approach and bonding by a proton, consequently, the electron pair which is present on N is unavailable for donation and hence 3º amines are the weakest bases.
Aromatic amines are weaker bases than ammonia and aliphatic amines. Electron-donating groups such as -CH3, -OCH3, etc. increase the basicity while electron-withdrawing substitutes such as -NO2, -CN, halogens, etc. decrease the basicity of amines. The effect of these substituents is more at p than at m-positions.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which one of the following is the strongest base in aqueous solution?
  1. Methyl amine.
  2. Tri methyl amine.
  3. Aniline.
  4. Dimethyl amine.
  1. Which order ofbasicity is correct?
  1. Aniline > m-toluidine > o-toluidine
  2. Aniline> o-toluidine > m-toluidine
  3. o-toluidine > aniline> m-toluidine
  4. o-toluidine < aniline < m-toluidine
  1. What is the decreasing order of basicity of primary, secondary and tertiary ethylamines and NH3?
  1. NH3 > C2H5NH2 > (C2H5)2NH > (C2H5)3N
  2. (C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3
  3. (C2H5)2NH > C2H5NH2> (C2H5)3N > NH3
  4. (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3
  1. The order of basic strength among the following amines in benzene solution is:
  1. CH3NH2 > (CH3)3N > (CH3)2NH
  2. (CH3)3N > (CH3)2NH > CH3NH2
  3. CH3NH2 > (CH3)2NH > (CH3)3N
  4. (CH3)3N > CH3NH2 > (CH3)2NH
  1. Choose the correct statement.
  1. Methylamine is slightly acidic.
  2. Methylamine is less basic than ammonia.
  3. Methylamine is a stronger base than ammonia.
  4. Methylamine forms salts with alkalies.
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Read the passage given below and answer the following questions:

Lucas test is a test to differentiate between primary, secondary and tertiary alcohols. This test consists of treating an alcohol with Lucas' reagent, and turbidity, due to the formation of insoluble alkyl chloride, is observed. Lucas test is based on the difference in reacting of three classes of alcohols with hydrogen chloride via SN1 reaction. The different reactivity reflects the differing ease of formation of the corresponding carbocations.

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: Equimolar mixture of cone. HCI and anhydrous ZnCl2 is called Lucas' reagent.

Reason: Lucas' reagent can be used to distinguish between methanol and ethanol.

  1. Assertion: 2-Methyl-2-butanol gives no turbidity with Lucas' reagent at room temperature.

Reason: It is a 3º alcohol.

  1. Assertion: Tertiary alcohols react fastest with Lucas' reagent by SN1 mechanism.

Reason: 3º carbocation is most stable.

  1. Assertion: Amongst the compounds, H2C = CHCH2OH (I), C6H5OH (II), CH3CH2CH2OH (III) and (CH3)3COH (IV), only (IV) reacts with Lucas' reagent at room temperature.

Reason: Tertiary alcohol gives turbidity immediately with Lucas' reagent.

  1. Assertion: Lucas test can be used to distinguish between 1-propanol and 2-propanol.

Reason: Lucas test is based upon the difference in reactivity of primary, secondary and tertiary alcohols with cone. HCI and anhyd. ZnCl2.

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Read the passage given below and answer the following questions:

Glucose is known as dextrose because it occurs in nature as the optically active dextrorotatory isomer. lt is essential constituent of human blood. The blood normally contains 65 to 110mg of glucose per 100mL (hence named Blood sugar). The level may be much higher in diabetic persons. The urine of diabetic persons also contain considerable amount of glucose. ln combined form, it occurs in cane sugar and polysaccharides such as starch and cellulose.

Glucose has an aldehyde group (-CHO), one primary alcoholic group (-CH2OH) and four secondary alcoholic groups (-CHOH) in their structure. Due to the presence five hydroxyl groups (-OH). glucose undergoes acetylation. Glucose also undergoes oxidation with mild oxidising agents like bromine water as well as with strong oxidising agents like nitric acid. Since glucose is readily oxidised, it acts as a strong reducing agent and reduces Tollen's reagent and Fehling solution. Glucose exists in two crystalline forms: $\alpha$-D-glucose and $\beta$-D-glucose. If either of the two forms is dissolved in water and allowed to stand, the specific rotation of the solution changes gradually, until a constant value is obtained. This change is called mutarotation.

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: A diabetic person carries a packet of glucose with him always.

Reason: Glucose increases the blood sugar level almost instantaneously.

  1. Assertion: On oxidation with nitric acid, glucose as well as gluconic acid both yield saccharic acid.

Reason: The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free -CHO group.

  1. Assertion: Glucose reacts with acetyl chloride to form pentaacetyl glucose.

Reason: The formation of pentaacetyl derivative confirms the presence of five -OH groups in glucose.

  1. Assertion: A certain compound gives negative test with ninhydrin and positive test with Benedict's solution, the compound is an amino acid.

Reason: Glucose is a monosaccharide.

  1. Assertion: The rapid interconversion of $\alpha$-D-glucose and $\beta$-D-glucose in solution is known as racemisation.

Reason: Hydrolysis reaction will take place when a mineral acid is treated with sugar.

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Read the passage given below and answer the following questions:

Carbohydrates can exist in either of two conformations, as determined by the orientation of the hydroxyl group about the asymmetric carbon farthest from the carbonyl.

By convention, a monosaccharide is said to have D-configuration if the hydroxyl group attached to the asymmetric carbon atom adjacent to the -CH2OH group is on the right hand side irrespective of the positions of the other hydroxyl groups. On the other hand, the molecule is assigned L-configuration if the -OH group attached to the carbon adjacent to the - CH2OH group is on the left hand side.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. D-Glyceraldehyde and L-Glyceraldehyde are:
  1. Epimers.
  2. Enantiomers.
  3. Anomers.
  4. Conformational diasteriomers.
  1. Which of the following monosaccharides, is the majority found in the human body?
  1. D-type.
  2. L-type.
  3. Both of these.
  4. None of these.
  1. The two functional groups present in a typical carbohydrate are:
  1. -OH and -COOH
  2. -CHO and -COOH
  3. >C= O and -OH
  4. -OH and -CHO
  1. Monosaccharides contain:
  1. Always six carbon atoms.
  2. Always five carbon atoms.
  3. Always four carbon atoms.
  4. May contain 3 to 7 carbon atoms.
  1. The correct corresponding order of names of four aldoses with configuration given below respectively, is:

  1. L-erythrose, L-threose, L-erythrose, D-threose.
  2. D-threose, D-erythrose, L-threose, L-erythrose.
  3. L-erythrose, L-threose, D-erythrose, D-threose.
  4. D-erythrose, D-threose, L-erythrose, L-threose.
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Read the passage given below and answer the following questions:

Carboxylic acids having an $\alpha$-hydrogen atom when treated with chlorine or bromine in the presence of small amount of red phosphorus gives $\alpha$-halocarboxytic acids. The reaction is known as Hell-Volhard-Zelinsky reaction.

$\text{R}-\text{CH}_2-\text{COOH}+\text{X}_2\xrightarrow{\text{red p}}\text{R}-\text{CH}-\text{COOH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{X}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{X = Cl, Br)}$

When sodium salt of carboxylic acid is heated with soda lime it loses carbon dioxide and gives hydrocarbon with less number of C-atoms.

$\text{R}-\text{COOH}\xrightarrow{\text{NaOH}}\text{R}-\text{COONa}\xrightarrow[\Delta]{\text{NaOH}+\text{CaO}}\text{R}-\text{H}+\text{Na}_2\text{CO}_3\\\text{Carboxylic}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Sod.}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Alkane}\\\ \ \ \ \ \text{acid}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{carboxylate}$

In these questions (Q. No. l-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: (CH3)3CCOOH does not give H.V.Z reaction.

Reason: (CH3)3CCOOH does not have $\alpha$-hydrogen atom.

  1. Assertion: H.V.Z. reaction involves the treatment of carboxylic acids having $\alpha$-hydrogens with Cl2 or Br2 in presence of small amount of red phosphorus.

Reason: Phosphorus reacts with halogens to form phosphorus trihalides.

  1. Assertion: Propionic acid with $\frac{\text{Br}_2}{\text{P}}$ yields CH2Br - CHBr - COOH.

Reason: Propionic acid has two $\alpha$-hydrogen atoms.

  1. Assertion: C6H5COCH2COOH undergoes decarboxylation easily than C6H5COCH2COOH.

Reason: C6H5COCH2COOH is $\beta$-keto acid.

  1. Assertion: On heating 3-methylbutanoic acid with soda lime, isobutane is obtained.

Reason: Soda lime is a mixture of NaOH + CaO in the ratio 3 : 1.

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Read the passage given below and answer the following questions:
The transition elements have incompletely filled d-subshells in their ground state or in any of their oxidation states. The transition elements occupy position in betweens- and p-blocks in groups 3-12 of the Periodic table. Starting from fourth period, transition elements consists of four complete series : Sc to Zn, Y to Cd and La, Hf to Hg and Ac, Rf to Cn. In general, the electronic configuration of outer orbitals of these elements is (n - 1)d1-10 n1-2. The electronic configurations of outer orbitals of Zn, Cd, Hg and Cn are represented by the general formula (n - 1)d10ns2. All the transition elements have typical metallic properties such as high tensile strength, ductility, malleability. Except mercury, which is liquid at room temperature, other transition elements have typical metallic structures. The transition metals and their compounds also exhibit catalytic property and paramagnetic behaviour. Transition metal also forms alloys. An alloy is a blend of metals prepared by mixing the components. Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which of the following characteristics of transition metals is associated with higher catalytic activity?
  1. High enthalpy of atomisation.
  2. Variable oxidation states.
  3. Paramagnetic behaviour.
  4. Colour of hydrated ions.
  1. Transition elements form alloys easily because they have.
  1. Same atomic number.
  2. Same electronic configuration.
  3. Nearly same atomic size.
  4. Same oxidation states.
  1. The electronic configuration of tantalum (Ta) is:
  1. [Xe]4f05d16s2
  2. [Xe]4f145d26s2
  3. [Xe]4f145d36s2
  4. [Xe]4f145d46s2
  1. Which one of the following outer orbital configurations may exhibit the largest number of oxidation states?
  1. 3d54s1
  2. 3d54s2
  3. 3d24s2
  4. 3d34s2
  1. The correct statement(s) among the following is/ are:
  1. All d and f-block elements are metals.
  2. All d and f-block elements form coloured ions.
  3. All d and f-block elements are paramagnetic.
  1. (I) only
  2. (I) and (II) only
  3. (II) and (III) only
  4. (I), (II) and (III)
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Read the passage given below and answer the following questions:

Fehling's reagent: Fehling's reagent is a mixture of two solutions. Fehllng's solution A is aqueous copper sulphate solution. Fehling's solution Bis alkaline sodium potassium tartarate (Rochelle salt).

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH(OH)COONa}\\\text{CuSo}_{4\text{(aq)}}+|\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH(OH)COOK}$

It is a mild oxidising agent. It is weaker than Tollens' reagent. It oxidises only aliphatic aldehydes to carboxylate ions and itself gets reduced to reddish brown precipitate of cuprous oxide. Aromatic aldehydes do not respond to Fehling's test. This reaction is used for the test of aliphatic aldehydes known as Fehling's reagent test.

In these questions (Q. No. l-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: Fehling's solution can be used to distinguish between acetaldehyde and acetone.

Reason: Fehling's reagent is a mixture of two solutions.

  1. Assertion: Aromatic aldehydes can be distinguished from aliphatic aldehydes by Fehling's solution.

Reason: Aromatic aldehydes reduce Fehling's solution, but aliphatic aldehydes do not.

  1. Assertion: Fehling's solution oxidises acetaldehyde to acetic acid but not benzaldehyde to benzoic acid.

Reason: The C-H bond of -CHO group in benzaldehyde is stronger than in acetaldehyde.

  1. Assertion: CH3CHO and C6H5CH2CHO cannot be distinguished chemically by Fehling's solution.

Reason: CH3CHO and C6H5CH2CHO cannot be distinguished chemically by Fehling's solution.

  1. Assertion: Formaldehyde, when heated with Fehling's reagent produces a reddish brown ppt, of Cu.

Reason: Fehling's reagent oxidises fonnaldehyde to formate ion.

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Read the passage given below and answer the following questions:
Nucleophilic substitution reactions are of two types; substitution nucleophilic bimolecular (SN2) and substitution nucleophilic unimolecular (SN1) depending on molecules taking part in determining the rate of reaction. Reactivity of alkyl halide towards SN1 and SN2 reactions depends on various factors such as steric hindrance, stability of intermediate or transition state, and polarity of solvent. SN2 reaction mechanism is favoured mostly by primary alkyl halide then secondary and then tertiary. This order is reversed in case of SN1 reactions.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which of the following is most reactive towards nucleophilic substitution reaction?
  1. C6H5Cl
  2. CH2 = CHCl
  3. ClCH2CH = CH2
  4. CH3CH = CHCl
  1. Isopropyl chloride undergoes hydrolysis by:
  1. SN1 mechanism.
  2. SN2 mechanism.
  3. SN1 and SN2 mechanism.
  4. Neither SN1 nor SN2 mechanism.
  1. The most reactive nucleophile among the following is:
  1. CH3O-
  2. C6H5O-
  3. (CH3)2CHO-
  4. (CH3)3CO-
  1. Tertiary alkyl halides are practically inert to substitution by SN2 mechanism because of:
  1. Insolubility.
  2. Instability.
  3. Inductive effect.
  4. Stearic hindrance.
  1. Which of the following is the correct order of decreasing SN2 reactivity?
  1. RCH2X > R2CHX > R3CX
  2. R3CX > R2CHX > RCH2X
  3. R2CHX > R3CX > RCH2X
  4. RCH2X > R3CX > R2CHX
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