14) Rama tested the solubility of four substances at different temperatures and found in grams, of each substance dissolved in 100 g of water to form a saturated solution.
Sl.no. Substance Dissolved Temperature (K)
(in grams) 293 K 313 K 333 K
(i) Ammonium chloride 37g 41 g 55g
(ii) Potassium chloride 35g 40g 46g
(iii) Sodium chloride 36g 36g 37g
(iv) Potassium Nitrate 32g 62g 106g
i) Which substance is least soluble in water at 293K?
ii) Which substance shows a maximum change in its solubility when the temperature is raised from 293K to 313K?
iii) Find the amount of ammonium chloride that will separate out when 155g of its solution at 333K is cooled to 293K?
iv) What is the effect of change of temperature on the solubility of a salt?
v) What mass of sodium chloride would be needed to make a saturated solution in 10g of water at 293K?
Answers
Answer:
(i) Potassium nitrate (1) (ii) Potassium nitrate (1) (iii) At 333K, ammonium chloride dissolved per 100g =55g At 293K, ammonium chloride dissolved perl00g=37g When solution is cooled from 333K to 293K, amount of ammonium chloride that will separate out =55g-37g=18g. (1) (iv) Solubility of a salt (soild) increases with rise in temperature and vice-versa. (1) (v) At 293K, amount of NaCI dissloved in 100g water = 36 g At 293K amount of NaCI dissolved in 10g water
Explanation:
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i) Potassium nitrate is least soluble in water at 293 K. <br> (ii) Potassium nitrate shows maximum change in its solubility which is 30 g (62-32) per 100 g of water. <br> (iii) From the available information, 55 g of ammonium chloride solution upon cooling from 333 K to 293 K separates salt = (55 - 37) = 18 g. <br> (iv) From the available information, it is clear that the solubility of salt in water increases with the rise in temperature <br> At 293 K, in a saturated solution, <br> 100 g of water has sodium chloride = 36 g <br> 10 g of water has sodium chloride = 3.6 g