1Revise, Reflect, Refine15 questions
Q.1Which of the following mixtures are correctly classified as homogeneous (Hm) and heterogeneous (Ht)? Choose the correct option. (i) Air — Hm, Milk — Ht, Sugar solution — Hm, Smoke — Hm (ii) Brass — Ht, Fog — Ht, Vinegar — Ht, Muddy water — Hm (iii) Copper sulfate solution — Hm, Salt solution — Hm, Milk — Hm, Bronze — Hm (iv) Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hmv
Answer:(iv) Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hm.
Q.2Choose the correct options, and explain the reason for the correct and incorrect options. Which among the following mixtures show the Tyndall Effect? A mixture of: (a) air and dust particles (b) copper sulfate and water (c) starch and water (d) acetone and water (i) a and b (ii) b and d (iii) a and c (iv) c and dv
Q.3A mixture can be categorised as a solution, a suspension, or a colloid, each possessing distinct properties. Utilise the words or phrases provided in the box to fill in the .2. Words and phrases may be used more than once. Words and Phrases Large-sized particles; Particles remain evenly distributed; Small-sized particles (less than 1 nm diameter); Moderate-sized particles (1 – 1000 nm); Settles down when left undisturbed (more than 1000 nm in diameter); Does not settle down; Scatters light; Separates by filtration; Transparent; Salt solution; Milk; Sand in water; Smoke; Heterogeneous mixture; Cannot be separated by filtration; Mud; Butter; Brass. .2. .2 Solution Suspension Colloid Properties Properties Properties __________________________ __________________________ __________________________ __________________________ __________________________ __________________________ Examples Examples Examples __________________________ __________________________ __________________________ __________________________ __________________________ __________________________v
Answer:Solution: homogeneous, particles less than 1 nm, transparent, do not settle, not separated by filtration; examples salt solution and brass. Suspension: heterogeneous, particles more than 1000 nm, settle on standing, separable by filtration; examples sand in water and mud. Colloid: heterogeneous, particles 1-1000 nm, scatter light, do not settle, not separated by ordinary filtration; examples milk, smoke and butter.
Q.4Solve the following problems: (i) g of sugar for 420 g of all-purpose flour and 5 g of sodium hydrogencarbonate. Express the concentration of each component in the mixture using an appropriate method. (ii) % copper by mass. Calculate the quantities of copper and zinc present in 120 g of brass.v
Answer:(i) Sugar = 15%, all-purpose flour = 84%, sodium hydrogencarbonate = 1% by mass. (ii) Copper = 84 g; zinc = 36 g.
Q.5The label on a cooking oil pack says one litre (910 g). If this oil is mixed with water, will it form a separate layer? If so, which substance will be on top? How will you separate the two layers? Also, draw the diagram of the apparatus used.v
Answer:Yes. Oil and water are immiscible and form separate layers. Oil is less dense, so it floats on top. Use a separating funnel: allow layers to settle, drain the lower water layer first, then collect the oil.
Q.6Assertion (A): Solutions do not exhibit the Tyndall effect. Reason (R): nm, so they cannot scatter light. Choose the correct option: (i) Both A and R are true, and R is the correct explanation of A. (ii) Both A and R are true, but R is not the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true.v
Answer:(iii) A is true, but R is false.
Q.7How would you separate the mixtures given in Table 5.3? Mention the reason for choosing your method. If a mixture cannot be separated, explain why. Mixtures: mud from muddy water; plasma from other components in the blood sample; naphthalene and sand; chalk powder and common salt; common salt and water; oil from water; pigments of the flower.v
Answer:Mud from muddy water—sedimentation/decantation and filtration; plasma from blood—centrifugation; naphthalene and sand—sublimation; chalk powder and common salt—dissolve salt, filter chalk, evaporate filtrate; common salt and water—evaporation/crystallisation; oil from water—separating funnel; flower pigments—chromatography.
Q.8Two miscible liquids, A and B, are present in a mixture. The boiling point of ° °C. Suggest a method to separate them. Also, draw a labelled diagram of the method suggested.v
Answer:Use distillation because the liquids are miscible and have different boiling points.
Q.9Compare evaporation, crystallization and distillation. In which situation, would you prefer each of these over the others?v
Answer:Evaporation removes solvent to leave solute and is useful when solvent recovery is not needed. Crystallization gives purer solid crystals from solution. Distillation vaporises and condenses liquid and is preferred for recovering a solvent or separating miscible liquids with different boiling points.
Q.10Blood is an example of a colloidal mixture. (i) What would happen if blood behaved like a true suspension inside the body? (ii) In a blood sample, identify the dispersed phase and the dispersion medium.v
Answer:If blood behaved like a true suspension, its cells/components would settle and circulation would be disrupted. In blood, the dispersed phase includes blood cells and colloidal particles; plasma is the dispersion medium.
Q.11You are given a mixture of sand, common salt and naphthalene (Fig. 5.25a). The Fig. 5.25b depicts various steps used to separate the components of this mixture. Identify and write down the correct sequence of separation techniques. Fig. 5.25: (a) 1 2 3 Fig. 5.25: (b)v
Answer:First sublime naphthalene from the mixture. Then add water to dissolve common salt, filter off sand, and evaporate or crystallise the filtrate to obtain common salt.
Q.12Why is distillation an effective method for separating a mixture of water and acetone?v
Answer:Distillation works because acetone and water are miscible liquids with different boiling points; acetone vaporises first and can be condensed separately.
Q.13Answer the following questions with the help of the data given in Table 5.4: Solubility of various salts (in g per 100 g of water) at different temperatures. (i) What mass of potassium nitrate would be needed to prepare its saturated solution in 50 g of water at 40 °C? (ii) A student makes a saturated solution of potassium chloride in water at 80 °C and leaves the solution to cool at room temperature (25 °C). What would she observe as the solution cools? Explain. (iii) What is the effect of a change in temperature on the solubility of salts? Also, compare the changes in the solubility of the four given salts with increasing temperature from 10 °C to 80 °C.v
Answer:At 40 °C, 62 g KNO₃ dissolves in 100 g water, so 31 g is needed for 50 g water. On cooling saturated KCl from 80 °C, crystals separate because solubility decreases. Solubility generally increases with temperature; KNO₃ increases sharply, NH₄Cl and KCl moderately, and NaCl very little.
Q.14Three students, A, B and C, are preparing sugar solutions for an experiment: y g of sugar in 80 g of water. y g of sugar in 100 g of water. y g of sugar in 80 g of water. (i) Calculate the mass percentage (% m/m) concentration of sugar in each student’s solution. (ii) Whose solution is the most concentrated? Explain why.v
Answer:Student A = 20%; Student B = 16.67%; Student C = 27.27%. Student C has the most concentrated solution.
Q.15Examine Fig. 5.26. (i) Identify the separation technique marked as ‘S’. (ii) Label the apparatus A, B and C. (iii) Which of the following mixtures can be separated by the technique identified above? Use the data B given in .5. Mixtures: A (a) water — acetone (b) water — salt (c) acetone — alcohol (d) sand — salt C (e) alcohol — chloroform (f) alcohol — benzene S ........................................... .5: Boiling points of some compounds Fig. 5.26 Solvent Water Acetone Alcohol Chloroform Benzene Temperature (°C) 100 ° ° ° ° °Cv
Answer:(i) The technique S is distillation (simple distillation). (ii) In the apparatus, A is the distillation flask that holds the mixture, B is the condenser where the vapour is cooled back to a liquid, and C is the receiver flask that collects the distillate. (iii) Simple distillation separates two miscible liquids whose boiling points differ widely. Using the boiling-point data (water 100 °C, acetone 56 °C, alcohol 78 °C, chloroform 61 °C, benzene 80 °C), the water–acetone mixture (a) is readily separated this way. Pairs with close boiling points — acetone–alcohol (c), alcohol–chloroform (e) and alcohol–benzene (f) — need fractional distillation instead, while water–salt (b) and sand–salt (d) are not liquid–liquid mixtures and are separated by evaporation or filtration rather than this setup.