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Chemistry Test - 15

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Chemistry Test - 15
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  • Question 1
    1 / -0

    The carbonyl stretching frequency for simple aldehydes, ketones, and carboxylic acids is about \(1710 \mathrm{~cm}^{-1}\), where the carbonyl stretching frequency for esters is about ................... \(\mathrm{cm}^{-1}\):

    Solution

    The carbonyl stretching frequency for esters is about1750 \(\mathrm{cm}^{-1}\)

    Carbonyl stretching peaks generally fall between 1900 and 1600 cm-1 (assume all peak positions hereafter are in wavenumber units), a relatively unique part of the IR spectrum. This area is sometimes referred to as the carbonyl stretching region as a result.Esters have a memorable pattern of three intense peaks at ~1700, ~1200, and ~1100 from the C=O. The peak at 1742 is of course the carbonyl stretch, and for saturated esters in general this peak falls from 1755 to 1735.

  • Question 2
    1 / -0

    Adenine is a derivative of __________.

    Solution

    Adenine is a derivative of purine. The nitrogenous bases are purines and pyrimidines. The purines are adenine and guanine. The pyrimidines are thymine and cytosine. The purines are double ring structure while the pyrimidines are the single ring made up of carbon and nitrogen. Purines pair with the pyrimidines by hydrogen bonds. The molecular formula is \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}_{5}\).

  • Question 3
    1 / -0

    For an ionic crystal of general formula AX and co-ordination number 6, the value of radius ratio will be:

    Solution

    Since the general formula is AX with coordination number 6, the crystal could be considered as \({NaCl}\). Thus, it is evident that the ratio lies between \(0.414\) and \(0.732\) because of its octahedral structure.

    The value ranging between \(0.414\) and \(0.732\) is obtained by dividing the radius of positive ion by that of negative ion.

  • Question 4
    1 / -0

    Positive Beilstein shows that:

    Solution

    Positive Beilstein shows that halogens may be present. A positive Beilstein’s test for halogens does not always indicate the presence of halogen since some halogen-free compounds viz. urea, thiourea, amides etc. also respond this test. The reason being the fact that these halogen-free compounds form cuprous cyanide which is volatile and decomposes to copper which burns with green flame.

  • Question 5
    1 / -0

    Calculate the amount of benzoic acid \(\left({C}_{6} {H}_{5} {COOH}\right)\) required for preparing 250 \({mL}\) of \(0.15 {M}\) solution in methanol.

    Solution

    0.15 M solution of benzoic acid in methanol means,

    1000 mL of solution contains 0.15 mol of benzoic acid.

    Therefore, \(250 {~mL}\) of solution contains \(\frac{0.15 \times 250}{1000}\) mol of benzoic acid

    \(=0.0375\) mol of benzoic acid

    Molar mass of benzoic acid \(\left({C}_{6} {H}_{5} {COOH}\right)=7 \times 12+6 \times 1+2 \times 16=122 {~g} {~mol}^{-1}\)

    Therefore, required benzoic acid \(=0.0375 {~mol} \times 122 {~g} {~mol}^{-1}=4.575 {~g}\)

  • Question 6
    1 / -0

    The process of passing of a precipitate into colloidal solution on adding an electrolyte is called:

    Solution

    The process of passing of a precipitate into colloidal solution on adding an electrolyte is calledPeptization.

    It is the process of converting is fresh precipitate into colloidal particles by shaking it with the dispersion medium in the presence of a small amount of suitable electrolyte called peptizing agent.

    Generally an electrolyle is added whose one ion is common with one ion of precipitate.The particles of the precipitate absorb common ion of the electrolyte. then they get dispersed due to electrostatic repulsion.

    Example: when \(\mathrm{Fe}Cl_{3}\) is added to \(\mathrm{Fe}(\mathrm{OH})_{3}\), the precipitate of hydromide changes to solution by absorbing \(Fe^{3+}\) Cons. \(\mathrm{Fe}^{3 +}\) is absorbed on \(\mathrm{Fe}(\mathrm{OH})_{3}\) and breaks into colloids.

  • Question 7
    1 / -0

    If a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C, calculate the osmotic pressure in Pascal exerted by it?

    Solution

    It is given that:

    Volume of water \(({V})=450 {~mL}=0.45 {~L}\)

    Temperature \((T)=37+273=310 {~K}\)

    Number of moles of the polymer,\(n=\frac{\text{Mass of polymer}}{\text{Molar Mass of polymer}}\)

    \(=\frac{1}{185000}\) mol

    We know that:

    Osmotic pressure, \(\pi=\frac{n}{V} R T\)

    where,\(R=8.314 \times 10^{3} {PaL} {K}^{-1} {~mol}^{-1}\)

    \(=\frac{1}{185000} {~mol} \times \frac{1}{0.45 L} \times 8.314 \times 10^{3} {PaL} {K}^{-1} {~mol}^{-1} \times 310 {~K}\)

    \(=30.98 {~Pa}\)

    \(=31 {~Pa}\)

  • Question 8
    1 / -0

    Alkanes can be prepared from Grignard reagents by reacting with:

    Solution

    Grignard reagents react rapidly with acidic hydrogen atoms in molecules such as alcohols and water, Primary amines, Alkynes, to produce alkanes.

    (A) Grignard reagents react rapidly with acidic hydrogen atoms in molecules such as alcohols and water. When a Grignard reagent reacts with water, a proton replaces the halogen, and the product is an alkane. The Grignard reagent therefore provides a pathway for converting a haloalkane to an alkane in two steps.

    (B) Though we typically want them to do nucleophilic addition to something, these powerful bases tend to go for protons -- a faster reaction -- if there is just about anything that might make the proton vulnerable. Attaching the proton to an electronegative atom like nitrogen is enough, so primary and even secondary amines will react like that with Grignard reagents. We just deprotonate the amine, the hydrocarbon part of the Grignard reagent ends up an alkane.

    (C) We recall that alkynes are more acidic than alkanes. Therefore, the acid–base reaction of an alkyne with a readily available Grignard reagent gives a Grignard reagent of the alkyne. This alkynide ion of the Grignard reagent reacts with carbonyl compounds.

  • Question 9
    1 / -0

    Which of the following is a linear molecule?

    Solution

    CO2is a linear molecule because of sp-hybridization around the carbon atom.

  • Question 10
    1 / -0

    One gram of charcoal adsorbs \(400 \mathrm{~mL}\) of \(0.5 \mathrm{M}\) acetic acid to form a monolayer, and the molarity of acetic acid reduces to \(0.49 \mathrm{M}\). Calculate the surface area of charcoal adsorbed by each molecule of acetic acid, where the surface area of charcoal is \(3.01 \times 10^{2} \mathrm{~m}^{2} \mathrm{~g}^{-1}\).

    Solution

    Given,

    One gram of charcoal adsorbs \(400 \mathrm{~mL}\) of \(0.5 \mathrm{M}\) acetic acid to form a monolayer, and the molarity of acetic acid reduces to \(0.49 \mathrm{M}\).

    \(400 \mathrm{~mL}\) of \((0.5-0.49=0.01) \mathrm{M}\) acetic acid contains \(0.4 \times\) \(0.01=0.2\) moles of acetic acid or \(4 \times 10^{-3} \times 6.023 \times 10^{23}=2.409 \times 10^{21}\) molecules of acetic acid

    Total surface area of \(1 \mathrm{~g}\) of charcoal is \(3.01 \times 10^{2} \mathrm{~m}^{2}\).

    Thus, the surface area occupied by 1 molecule of acetic acid \(=\frac{3.01 \times 10^{2} \mathrm{~m}^{2}}{2.409 \times 10^{21}}\)

    \(=1.2 \times 10^{-19} \mathrm{~m}^{2}\)

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