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Please solve A B and C. Show each step to the solutions. I attached the answers for B but am not sure how to get to these solutions.
please solve A B and C. Show each step to the solutions 6.82. A salt A is soluble in a solvent S. A conductivity meter used t
Problem 11 6.82 Answer: b) Solubility at 10.2°C= 41.0 g A/100 g S; Solubility at 0°C 11.8 g A/100 g S; Mass of solid A = 114
please solve A B and C. Show each step to the solutions 6.82. A salt A is soluble in a solvent S. A conductivity meter used to measure the solute concentration in A-S solutions is calibrated by dissolving a known quantity of A in S, adding more S to bring the solution volume to a fixed value, and noting the conductivity meter reading. The data given below are taken at 30°C: (b) Meter Solution Solute Reading Volume Dissolved R (mL) (g) 0 100.0 0 30 100.0 20.0 45 100.0 30.0 The following experiment is performed. One hundred sixty grams of A is dissolved in S at 30°C. S is added until a final solution volume of 500 mL is obtained. The solution is cooled slowly to 0°C while being stirred and is maintained at this temperature long enough for crystallization to be complete. The concentration of A in the supernatant liquid is then measured with the conductivity meter, yielding R= 17.5. The solution is next reheated in small temperature increments. The last crystal is observed to dissolve at 10.2°C. A specific gravity of 1.10 may be assumed for all A-S solutions. (a) Derive an expression for C(g A/mL solution) in terms of R. (b) Calculate the solubilities (g A/100 g S) at 10.2°C and 0°C and the mass of solid crystals in the beaker at 0°C. If half the solvent in the flask were to evaporate at 0°C, how much more A would come out of solution? (c)
Problem 11 6.82 Answer: b) Solubility at 10.2°C= 41.0 g A/100 g S; Solubility at 0°C 11.8 g A/100 g S; Mass of solid A = 114 g
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