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NCERT Exemplar · Class 9 Science Is Matter Around Us Pure

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Long Answer Questions 33–42 (part 4 of 4)

  1. Exercise 33

    Fractional distillation is suitable for separation of miscible liquids with a boiling point difference of about 25\displaystyle 25 K or less. What part of fractional distillation apparatus makes it efficient and possess an advantage over a simple distillation process. Explain using a diagram.

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    NCERT’s answer
    The fractionating column packed with glass beads provides a surface for the vapours to collide and lose energy so that they can be quickly condensed and distilled. Also length of the column would increase the efficiency. Figure: Fractional distillation
    The fractionating column, fitted between the flask and the condenser, is the part that makes the process efficient. NCERT_Solution_Class9_Science_Exemplar_Ch2_Q33 It is packed with glass beads that give the rising vapour a large surface on which to cool, condense and reflux repeatedly before escaping upward: \[\text{vapour (mixed)} \xrightarrow{\text{repeated condense-reflux on beads}} \text{vapour enriched in the lower-boiling component} \] Each cycle raises the vapour's share of the more volatile liquid, so components only \(\displaystyle 25\text{ K}\) apart in boiling point separate cleanly -- something a bead-free simple-distillation flask cannot do. Answer: The fractionating column — packed with beads for repeated condense-reflux cycles — is the part that makes fractional distillation efficient over simple distillation.

  2. Exercise 34

    (a)
    Under which category of mixtures will you classify alloys and why?
    (b)
    A solution is always a liquid. Comment.
    (c)
    Can a solution be heterogeneous?

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    NCERT’s answer
    Hint- (a) Homogenous mixture, because they have a uniform composition throughout (b) No, solid solutions and gaseous solutions are also possible. Examples brass and air (c) No, solution is a homogenous mixture of two or more substances
    (a)
    Homogeneous mixture -- an alloy is a solid solution of a metal with another metal or a non-metal, uniform composition throughout, e.g. brass (\(\displaystyle \text{Cu} + \text{Zn}\)), steel (\(\displaystyle \text{Fe} + \text{C}\)).
    (b)
    False. A solution's solvent and solute can each be solid, liquid or gas -- brass is a solid solution, soda water a gas-in-liquid solution, air a gas-in-gas solution.
    (c)
    No -- a solution is a homogeneous mixture by definition, so it has no separate phase to see.
    Answer: (a) Homogeneous mixture (metal + metal or non-metal). (b) False -- a solution need not be liquid. (c) No, a solution is always homogeneous.

  3. Exercise 35

    Iron filings and sulphur were mixed together and divided into two parts, ‘A’ and ‘B’. Part ‘A’ was heated strongly while Part ‘B’ was not heated. Dilute hydrochloric acid was added to both the Parts and evolution of gas was seen in both the cases. How will you identify the gases evolved?

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    NCERT’s answer
    Part A Fe (s) + S (s) Heat FeS (s) FeS + $\displaystyle 2$ HCl (aq) → \(\displaystyle FeCl_{2}\) + \(\displaystyle H_{2}\) S Part B Fe (s) + S (s)→ Mixture of iron filings and sulphur When dilute HCl is added to it Fe (s) + S (s) + $\displaystyle 2$ HCl (aq) → \(\displaystyle FeCl_{2}\) + \(\displaystyle H_{2}\) gas Sulphur remains unreacted \(\displaystyle H_{2}\)S gas formed has a foul smell and on passing through lead acetate solution, it turns the solution black. Hydrogen gas burns with a pop sound.
    Part B (unheated) is still a mechanical mixture of iron and sulphur; only the iron reacts with the acid, sulphur does not: \[\text{Fe(s)} + 2\text{HCl(aq)} \rightarrow \text{FeCl}_2\text{(aq)} + \text{H}_2\text{(g)} \] Part A was heated, so the iron and sulphur had already combined into one compound, iron(II) sulphide: \[\text{Fe(s)} + \text{S(s)} \xrightarrow{\Delta} \text{FeS(s)} \] \[\text{FeS(s)} + 2\text{HCl(aq)} \rightarrow \text{FeCl}_2\text{(aq)} + \text{H}_2\text{S(g)} \] The two gases are told apart by a burning splint and by smell: the gas from B, \(\displaystyle \text{H}_2\), pops with a lighted splint and is odourless; the gas from A, \(\displaystyle \text{H}_2\text{S}\), smells of rotten eggs and blackens moist lead-acetate paper, \[\text{Pb(CH}_3\text{COO)}_2 + \text{H}_2\text{S} \rightarrow \text{PbS(black)} + 2\text{CH}_3\text{COOH} \] Answer: Gas from B (unheated) is \(\displaystyle \text{H}_2\) — pops with a lit splint, odourless. Gas from A (heated) is \(\displaystyle \text{H}_2\text{S}\) — smells of rotten eggs and blackens moist lead-acetate paper.

  4. Exercise 36

    A child wanted to separate the mixture of dyes constituting a
    sample of ink. He marked a line by the ink on the filter paper and
    placed the filter paper in a glass containing water as shown in
    Fig.2.3. The filter paper was removed when the water moved near
    the top of the filter paper.
    (i)
    What would you expect to see, if the ink contains three different
    coloured components?
    (ii)
    Name the technique used by the child.
    (iii)
    Suggest one more application of this technique.
    NCERT_Question_Class9_Science_Exemplar_Ch2_Q36

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    NCERT’s answer
    Hint- (i) Three different bands will be observed. (ii) Chromatography (iii) To separate the pigments present in Chlorophyll.
    (i)
    The dyes travel up the wet paper at different rates set by their solubility in water and their adsorption onto the paper: \(\displaystyle \text{more soluble, less adsorbed} \Rightarrow \text{rises farther}\). Three separate coloured spots/bands would appear at different heights instead of one mixed colour.
    (ii)
    Paper chromatography.
    (iii)
    The same technique separates the pigments of a leaf extract (chlorophylls and carotenoids) into distinct coloured bands.
    Answer: (i) three separate coloured bands at different heights (ii) paper chromatography (iii) separating leaf (chlorophyll) pigments.

  5. Exercise 37

    A group of students took an old shoe box
    and covered it with a black paper from all
    sides. They fixed a source of light (a torch)
    at one end of the box by making a hole in it
    and made another hole on the other side to
    view the light. They placed a milk sample
    contained in a beaker/tumbler in the box
    as shown in the Fig.2.4. They were amazed
    to see that milk taken in the tumbler was
    illuminated. They tried the same activity by taking a salt solution but found
    that light simply passed through it?
    (a)
    Explain why the milk sample was illuminated. Name the phenomenon
    involved.
    (b)
    Same results were not observed with a salt solution. Explain.
    (c)
    Can you suggest two more solutions which would show the same effect
    as shown by the milk solution?
    NCERT_Question_Class9_Science_Exemplar_Ch2_Q37

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    NCERT’s answer
    (a)
    Milk is a colloid and would show Tyndall effect. (b) Salt solution is a true solution and would not scatter light. (c) Detergent solution, sulphur solution.
    (a)
    Milk is a colloid: \(\displaystyle \text{colloidal particles} \Rightarrow \text{scatter light} \Rightarrow \text{beam visible sideways}\) — the Tyndall effect.
    (b)
    A salt solution is a true solution: the dissolved \(\displaystyle \text{Na}^+\) and \(\displaystyle \text{Cl}^-\) ions are far smaller than light's wavelength, too small to scatter it appreciably, so the beam passes through unseen from the side.
    (c)
    Any other colloid shows it too, e.g. a starch solution or a soap solution.
    Answer: (a) colloidal particles scatter light: Tyndall effect (b) true solution, particles too small to scatter light (c) starch solution, soap solution.

  6. Exercise 38

    Classify each of the following, as a physical or a chemical change. Give
    reasons.
    (a)
    Drying of a shirt in the sun.
    (b)
    Rising of hot air over a radiator.
    (c)
    Burning of kerosene in a lantern.
    (d)
    Change in the colour of black tea on adding lemon juice to it.
    (e)
    Churning of milk cream to get butter.

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    NCERT’s answer
    Hint—Physical changes —(a), (b), (e) Chemical changes— (c), (d)
    (a)
    Physical change -- only a state change (liquid \(\displaystyle \rightarrow\) vapour); the water can be recovered, no new substance forms.
    (b)
    Physical change -- heat lowers the air's density, so it rises; the air's composition is unchanged.
    (c)
    Chemical change -- kerosene burns in oxygen to \(\displaystyle \text{CO}_2\) and \(\displaystyle \text{H}_2\text{O}\), new substances that release heat and light.
    (d)
    Chemical change -- acid protonates tea's coloured tannins, turning them into a different, lighter-coloured compound.
    (e)
    Physical change -- churning only coalesces the same fat globules into a lump; no new substance forms.
    Answer: (a) physical (b) physical (c) chemical (d) chemical (e) physical.

  7. Exercise 39

    During an experiment the students were asked to prepare a 10\displaystyle 10% (Mass/Mass)
    solution of sugar in water. Ramesh dissolved 10g of sugar in 100g of water
    while Sarika prepared it by dissolving 10g of sugar in water to make 100g
    of the solution.
    (a)
    Are the two solutions of the same concentration
    (b)
    Compare the mass % of the two solutions.

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    NCERT’s answer
    (a)
    No. Mass % = Massof solute ×$\displaystyle 100$ Mass of solute + Mass of solvent (b) Solution made by Ramesh Mass % = $\displaystyle 100$ = $\displaystyle 10$ +$\displaystyle 100$ $\displaystyle 10$ ×$\displaystyle 100$ = $\displaystyle 9.09$% Solution made by Sarika Mass % = ×$\displaystyle 100$ = $\displaystyle 10$% The solution prepared by Sarika has a higher mass % than that prepared by Ramesh.
    Ramesh's solution has \(\displaystyle 10\text{ g}\) sugar added to \(\displaystyle 100\text{ g}\) water, giving a total mass of \(\displaystyle 110\text{ g}\):
    \[\text{Mass \%} = \frac{\text{mass of sugar}}{\text{mass of solution}}\times100 = \frac{10}{10+100}\times100 = 9.09\% \]
    Sarika's \(\displaystyle 10\text{ g}\) sugar makes up the whole \(\displaystyle 100\text{ g}\) of solution directly:
    \[\text{Mass \%} = \frac{10}{100}\times100 = 10\% \]
    (a)
    The two are not the same concentration.
    (b)
    Sarika's solution, at \(\displaystyle 10\%\), is slightly more concentrated than Ramesh's \(\displaystyle 9.09\%\).
    Answer: (a) no, the two solutions are not the same concentration (b) Sarika's \(\displaystyle 10\%\) is more concentrated than Ramesh's \(\displaystyle 9.09\%\).

  8. Exercise 40

    You are provided with a mixture containing sand, iron filings, ammonium chloride and sodium chloride. Describe the procedures you would use to separate these constituents from the mixture?

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    NCERT’s answer
    Hint- Step-$\displaystyle 1$ Separate iron filings with the help of a magnet Step-$\displaystyle 2$ Sublimation of the remaining mixture separates ammonium chloride Step-$\displaystyle 3$ Add water to the remaining mixture, stir and filter Step-$\displaystyle 4$ The filtrate can be evaporated to get back sodium chloride.
    A magnet passed through the mixture pulls out the iron filings -- magnetic separation. The remainder (sand, \(\displaystyle \text{NH}_4\text{Cl}\), \(\displaystyle \text{NaCl}\)) is heated in a china dish under an inverted funnel with a cotton-plugged stem, subliming the ammonium chloride BEFORE water is added, since \(\displaystyle \text{NH}_4\text{Cl}\) is water-soluble and would otherwise dissolve into the filtrate with the \(\displaystyle \text{NaCl}\): \[\text{NH}_4\text{Cl(s)} \xrightarrow{\Delta} \text{NH}_3\text{(g)} + \text{HCl(g)} \xrightarrow{\text{cool funnel wall}} \text{NH}_4\text{Cl(s)} \]Heat gently -- a fierce flame drives vapour past the plug before it condenses. Sand and \(\displaystyle \text{NaCl}\) remain; add water, stir, and filter. Sand is the residue; \(\displaystyle \text{NaCl}\) passes into the filtrate. Evaporate it gently, stopping short of dryness, to recover \(\displaystyle \text{NaCl}\) crystals without spattering.Answer: Fe by magnet; \(\displaystyle \text{NH}_4\text{Cl}\) by sublimation before water is added; sand as residue on filtering; \(\displaystyle \text{NaCl}\) by evaporating the filtrate.

  9. Exercise 41

    Arun has prepared 0.01\displaystyle 0.01% (by mass) solution of sodium chloride in water.
    Which of the following correctly represents the composition of the solutions?
    (a)
    1.00\displaystyle 00 g of NaCl + 100g of water
    (b)
    0.11g of NaCl + 100g of water
    (c)
    0.01\displaystyle 01 g of NaCl + 99.99g of water
    (d)
    0.10\displaystyle 10 g of NaCl + 99.90g of water

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    NCERT’s answer
    (c)
    Mass % = Massof solute Massof solute Massof solvent × + = $\displaystyle 0.01$ $\displaystyle 0.01$ $\displaystyle 99.99$ × + = $\displaystyle 0.01$ $\displaystyle 100$ × = $\displaystyle 0.01$ g
    (c) \(\displaystyle 0.01\ \text{g NaCl} + 99.99\ \text{g water}\) \[\text{Mass \%} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100 = \frac{0.01\ \text{g}}{(0.01 + 99.99)\ \text{g}} \times 100 = 0.01\% \] The others give $\displaystyle 0.99$%, $\displaystyle 0.11$% and $\displaystyle 0.10$%.Answer: (c) \(\displaystyle 0.01\ \text{g NaCl} + 99.99\ \text{g water}\).

  10. Exercise 42

    Calculate the mass of sodium sulphate required to prepare its 20\displaystyle 20% (mass percent) solution in 100g of water?

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    NCERT’s answer
    Let the mass of sodium sulphate required be = x g The mass of solution would be = (x +$\displaystyle 100$) g x g of solute in (x + $\displaystyle 100$) g of solution $\displaystyle 20$% = ×$\displaystyle 100$ +$\displaystyle 100$ x x $\displaystyle 20$ x + $\displaystyle 2000$= $\displaystyle 100$ x $\displaystyle 80$ x = $\displaystyle 2000$ x = $\displaystyle 2000$ = $\displaystyle 25$ g
    \[m_{\text{solvent}} = 100\ \text{g}, \quad m_{\text{solute}} = x\ \text{g} \] \[\text{Mass \%} = \frac{m_{\text{solute}}}{m_{\text{solute}} + m_{\text{solvent}}} \times 100 \] \[20 = \frac{x\ \text{g}}{(x + 100)\ \text{g}} \times 100 \] \[20(x + 100) = 100x \] \[2000 = 80x \] \[x = 25\ \text{g} \] Answer: $\displaystyle 25$ g of \(\displaystyle \text{Na}_2\text{SO}_4\).