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NCERT Exemplar · Class 9 Science Is Matter Around Us Pure

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Short Answer Questions 10–19 (part 2 of 4)

  1. Exercise 10

    Suggest separation technique(s) one would need to employ to
    separate the following mixtures.
    (a)
    Mercury and water
    (b)
    Potassium chloride and ammonium chloride
    (c)
    Common salt, water and sand
    (d)
    Kerosene oil, water and salt

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    NCERT’s answer
    (a)
    Separation by using separating funnel (b) Sublimation (c) Filtration followed by evaporation or Centrifugation followed by evaporation/distillation (d) Separation by using separating funnel to separate kerosene oil followed by evaporation or distillation.
    (a)
    Immiscible, very different densities \(\displaystyle \Rightarrow\) separating funnel.
    (b)
    On heating, \(\displaystyle \text{NH}_4\text{Cl}\) sublimes while \(\displaystyle \text{KCl}\) does not \(\displaystyle \Rightarrow\) sublimation separates them.
    (c)
    Sand is insoluble in water \(\displaystyle \Rightarrow\) filtration removes it; the filtrate is then evaporated to recover the salt.
    (d)
    Immiscible: the denser salt solution is drained from under the kerosene, then evaporated.
    Answer: (a) separating funnel (b) sublimation (c) filtration, then evaporation (d) separating funnel, then evaporation.

  2. Exercise 11

    Which of the tubes in Fig. 2.1\displaystyle 2.1 (a) and (b) will be more effective as a condenser in the distillation apparatus? NCERT_Question_Class9_Science_Exemplar_Ch2_Q11

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    NCERT’s answer
    Hint— Look for the larger surface area. The presence of beads in tube (a) would provide a larger surface area for cooling.
    Tube (a) is packed with glass beads, which give it a far larger cooled glass surface than the plain tube (b): \(\displaystyle \text{beads}\Rightarrow\text{surface area}\uparrow\Rightarrow\text{cooling}\uparrow\). Over the same length, the vapour touches more cooled surface and loses heat, i.e. condenses, faster.Answer: Tube (a); its glass beads give a larger cooling surface, so the vapour condenses faster.

  3. Exercise 12

    Salt can be recovered from its solution by evaporation. Suggest some other technique for the same?

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    NCERT’s answer
    Crystallization
    Crystallisation. The solution is heated gently to concentrate it, stopping well short of dryness so soluble impurities stay dissolved, then cooled: pure \(\displaystyle \text{NaCl}\) crystals separate out and are removed by filtration. Heating fully to dryness would crystallise the impurities too.Answer: Crystallisation.

  4. Exercise 13

    The ‘sea-water’ can be classified as a homogeneous as well as heterogeneous mixture. Comment.

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    NCERT’s answer
    Homogeneous— mixture of salts and water only Heterogeneous— contains salts, water, mud, decayed plant etc.
    Sea water is a solution of dissolved salts (e.g. \(\displaystyle \text{NaCl}\), \(\displaystyle \text{MgCl}_2\)) in water — one phase throughout, so it is homogeneous. Natural sea water also carries suspended sand, silt and marine debris, forming visible separate phases — making that same sample heterogeneous too.Answer: homogeneous as a salt solution; heterogeneous once suspended sand, silt and debris are included.

  5. Exercise 14

    While diluting a solution of salt in water, a student by mistake added acetone (boiling point 56\displaystyle 56°C). What technique can be employed to get back the acetone? Justify your choice.

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    NCERT’s answer
    Hint— Distillation, since acetone is more volatile it will separate out first.
    Acetone (bp \(\displaystyle 56^\circ\text{C}\)) is fully miscible with water, so a separating funnel cannot split the layers; its boiling point is far below that of the salt solution (bp \(\displaystyle >100^\circ\text{C}\)), so distillation is used instead. Heating gently on a water bath, acetone vaporises first and is condensed and collected, leaving the salt solution behind — direct strong heating is avoided since acetone is highly volatile and flammable.Answer: distillation on a water bath.

  6. Exercise 15

    What would you observe when
    (a)
    a saturated solution of potassium chloride prepared at 60\displaystyle 60°C is allowed
    to cool to room temperature.
    (b)
    an aqueous sugar solution is heated to dryness.
    (c)
    a mixture of iron filings and sulphur powder is heated strongly.

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    NCERT’s answer
    (a)
    Solid potassium chloride will separate out. (b) Initially the water will evaporate and then sugar will get charred. (c) Iron sulphide will be formed.
    (a)
    Cooling lowers KCl's solubility, so the dissolved excess crystallises out.
    (b)
    Water leaves first; on strong heating the dry sugar chars and blackens (never heat to dryness).
    (c)
    Iron and sulphur combine on heating:
    \[\text{Fe} + \text{S} \xrightarrow{\Delta} \text{FeS} \]
    the grey-black compound no longer attracted by a magnet.
    Answer: (a) KCl crystals separate out. (b) Sugar chars and blackens. (c) Grey-black FeS forms, non-magnetic.

  7. Exercise 16

    Explain why particles of a colloidal solution do not settle down when left undisturbed, while in the case of a suspension they do.

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    NCERT’s answer
    Particle size in a suspension is larger than those in a colloidal solution. Also molecular interaction in a suspension is not strong enough to keep the particles suspended and hence they settle down.
    Colloidal particles are extremely small and light, so they are struck unevenly and continuously by fast-moving medium molecules, producing a random zig-zag \(\displaystyle \left(\text{Brownian}\right)\) motion that keeps them suspended against gravity. Suspension particles are far larger and heavier, so gravity dominates over these collisions and they settle under gravity. Answer: Continuous, unequal molecular bombardment keeps colloidal particles in Brownian motion so they never settle; suspension particles are too large and heavy for this and settle under gravity.

  8. Exercise 17

    Smoke and fog both are aerosols. In what way are they different?

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    NCERT’s answer
    Both fog and smoke have gas as the dispersion medium. The only difference is that the dispersed phase in fog is liquid and in smoke it is a solid
    Fog is a liquid dispersed in a gas — tiny water droplets suspended in air. Smoke is a solid dispersed in a gas — fine carbon/ash particles suspended in air. Both share air as the dispersion medium; only the dispersed phase differs. Answer: Fog is liquid dispersed in gas (water droplets in air); smoke is solid dispersed in gas (carbon/ash particles in air) — same medium, different dispersed phase.

  9. Exercise 18

    Classify the following as physical or chemical properties
    (a)
    The composition of a sample of steel is: 98\displaystyle 98% iron, 1.5\displaystyle 1.5% carbon and
    0.5\displaystyle 5% other elements.
    (b)
    Zinc dissolves in hydrochloric acid with the evolution of hydrogen gas.
    (c)
    Metallic sodium is soft enough to be cut with a knife.
    (d)
    Most metal oxides form alkalis on interacting with water.

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    NCERT’s answer
    Physical properties - (a) and (c) Chemical properties - (b) and (d)
    (a)
    Physical — composition is measured without changing the steel into a new substance.
    (b)
    Chemical — a new substance and a gas appear:
    \[\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g) \]
    (c)
    Physical — softness (ease of cutting) is observed without changing sodium into anything else.
    (d)
    Chemical — the oxide reacts with water to give a new substance:
    \[\text{Metal oxide} + \text{H}_2\text{O} \rightarrow \text{Metal hydroxide (alkali)} \]
    Answer: Physical: (a), (c); Chemical: (b), (d).

  10. Exercise 19

    The teacher instructed three students ‘A’, ‘B’ and ‘C’ respectively to prepare a 50\displaystyle 50% (mass by volume) solution of sodium hydroxide (NaOH). ‘A’ dissolved 50g of NaOH in 100\displaystyle 100 mL of water, ‘B’ dissolved 50g of NaOH in 100g of water while ‘C’ dissolved 50g of NaOH in water to make 100\displaystyle 100 mL of solution. Which one of them has made the desired solution and why?

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    NCERT’s answer
    ‘C’ has made the desired solution Mass by volume % = Mass of solute ×$\displaystyle 100$ Volume of solution = $\displaystyle 50$ ×$\displaystyle 100$ = $\displaystyle 50$ % mass by volume
    Mass by volume percentage is defined per $\displaystyle 100$ mL of solution: \[\%(w/V) = \frac{\text{mass of solute}}{\text{volume of solution}} \times 100 \] Only C dissolves \(\displaystyle 50\text{ g}\) \(\displaystyle \text{NaOH}\) in water and then makes the total solution volume up to \(\displaystyle 100\text{ mL}\): \[\%(w/V) = \frac{50\text{ g}}{100\text{ mL}} \times 100 = 50\% \] A uses \(\displaystyle 100\text{ mL}\) of water, not of solution, so the final volume exceeds \(\displaystyle 100\text{ mL}\); B uses mass of water, giving a mass/mass percentage, not mass/volume. Answer: C — only C dissolves $\displaystyle 50$ g NaOH in water and makes the solution up to $\displaystyle 100$ mL, giving the required $\displaystyle 50$% mass/volume.