Answer the question in the attached images please. Solubility [g/(100. g H,O)]Substance20 °C50 CKCI3443NaNO311088C12H22O11 (sugar)204260You may want to reference (Pages 294 - 298) Section 9.3 whilecompleting this problem. Part ADetermine whether each of the following solutions will be saturated or unsaturated at 50 °C:Drag the appropriate items to their respective bins.ResetHelpadding 23 g of KCIto 50. g of H,Oadding 50. g of sugarto 25. g of H2Oadding 160. g of NaNO,to 75. g of H2OSaturated-UnsaturatedSubrnitRequest Answerbp

The data of solubility of each compenent is given for 100 g of water at 50⁰C For KCl, solubility at 50⁰C= 43 g per 100 g water For sugar, solubility at 50⁰C= 260 g per 100 g water For NaNO3, solubility at 50⁰C= 110 g per 100 g water (a) Now, given that 23 g of KCl is added in 50 g water. Since solubility of KCl = 43/ 100 = 0.43 g for each gram of water Hence, for 50 g of water , solubility of KCl = 50 ×0.43 =21.5 g Since, the taken amount of KCl is 23 g which is greater than 21.5 g. Hence, it is saturated solution. (b) Given that 50 g of sugar is added in 25 g water. Since solubility of sugar = 260/ 100 = 2.6 g for each gram of water Hence, for 25 g of water , solubility of sugar = 25 ×2.6 =65 g Since, the taken amount of sugar is 50 g only which is less than 65 g. Hence, it is unsaturated solution.(c) Now, given that160 g of NaNO3 is added in 75 g water. Since solubility of NaNO3= 110/ 100 = 1.1 g for each gram of water Hence, for 75 g of water , solubility of NaNO3= 75 ×1.1 =82.5 g Since, the taken amount of NaNO3 is160 g which is greater than require 82.5 g. Hence, it is saturated solution.