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Three students, A, B and C, are preparing sugar solutions for an experiment: Student A dissolves 20 g of sugar in 80 g of water. Student B dissolves 20 g of sugar in 100 g of water. Student C dissolves 30 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.
For Student A, total solution mass is 100 g, so concentration is 20/100 × 100 = 20 percent. For Student B, it is 20/120 × 100 = 16.67 percent. For Student C, it is 30/110 × 100 = 27.27 percent. Therefore, Student C has the most concentrated solution because the percentage of sugar in the total solutRead more
For Student A, total solution mass is 100 g, so concentration is 20/100 × 100 = 20 percent. For Student B, it is 20/120 × 100 = 16.67 percent. For Student C, it is 30/110 × 100 = 27.27 percent. Therefore, Student C has the most concentrated solution because the percentage of sugar in the total solution is highest.
For more NCERT Solutions of Class 9 Science Exploration Chapter 5 Exploring Mixtures and their Separation Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-5/
See lessAnswer the following questions with the help of the data given in Table 5.4. (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.
(i) At 40 °C, potassium nitrate has solubility of 62 g per 100 g water. Therefore, 50 g water needs 31 g potassium nitrate. (ii) Potassium chloride will crystallise as cooling decreases its solubility. (iii) From 10 °C to 80 °C, potassium nitrate increases from 21 to 167 g, sodium chloride from 36 tRead more
(i) At 40 °C, potassium nitrate has solubility of 62 g per 100 g water. Therefore, 50 g water needs 31 g potassium nitrate. (ii) Potassium chloride will crystallise as cooling decreases its solubility. (iii) From 10 °C to 80 °C, potassium nitrate increases from 21 to 167 g, sodium chloride from 36 to 37 g, potassium chloride from 35 to 54 g and ammonium chloride from 24 to 66 g.
For more NCERT Solutions of Class 9 Science Exploration Chapter 5 Exploring Mixtures and their Separation Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-5/
See lessWhy is distillation an effective method for separating a mixture of water and acetone?
Distillation effectively separates water and acetone because their boiling points differ considerably. Acetone boils at about 56 °C, whereas water boils at 100 °C. On heating, acetone vaporises before water forms significant vapour. The acetone vapour then passes through the condenser, where it coolRead more
Distillation effectively separates water and acetone because their boiling points differ considerably. Acetone boils at about 56 °C, whereas water boils at 100 °C. On heating, acetone vaporises before water forms significant vapour. The acetone vapour then passes through the condenser, where it cools and changes back into liquid. Thus, the two liquids can be separated effectively using distillation because their boiling points differ sufficiently.
For more NCERT Solutions of Class 9 Science Exploration Chapter 5 Exploring Mixtures and their Separation Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-5/
See lessYou 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.
The correct sequence is 1, 3, 2. Step 1 uses sublimation to separate naphthalene because it changes directly into vapour on heating. The remaining sand and common salt are treated with water to dissolve the salt. Step 3 uses filtration to separate the insoluble sand. Finally, step 2 uses evaporationRead more
The correct sequence is 1, 3, 2. Step 1 uses sublimation to separate naphthalene because it changes directly into vapour on heating. The remaining sand and common salt are treated with water to dissolve the salt. Step 3 uses filtration to separate the insoluble sand. Finally, step 2 uses evaporation to remove water and obtain common salt as the remaining solid.
For more NCERT Solutions of Class 9 Science Exploration Chapter 5 Exploring Mixtures and their Separation Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-5/
See lessBlood 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.
If blood behaved like a true suspension, its larger particles would settle down when the blood was left undisturbed, rather than remaining dispersed. Blood is actually a colloid. In a blood sample, the dispersed phase consists of the dispersed blood cells and other particles, while plasma acts as thRead more
If blood behaved like a true suspension, its larger particles would settle down when the blood was left undisturbed, rather than remaining dispersed. Blood is actually a colloid. In a blood sample, the dispersed phase consists of the dispersed blood cells and other particles, while plasma acts as the dispersion medium in which these particles are suspended. Centrifugation can separate blood components based on their properties.
For more NCERT Solutions of Class 9 Science Exploration Chapter 5 Exploring Mixtures and their Separation Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-5/
See less