Question

13.5-2. One hundred grams of a particular solute are dissolved in one liter of water. The vapor pressure of the water is decreased by roughly 6%. Is the solute more likely to be sugar (C2H201), table salt (NaCI), or potassium iodide (KI)? Simple ionic solutions double their effective Raoult concentration!

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Answer 1

From Raoult's P_0 - P/P_0 = x_2 where P_0 = pressure of pure P = pressure of solution x_2 = mol fraction of solute. Now, 100g of solute is dissolved in 1L assuming has density of 1g/mL. (Molar mass of NaCl = 58.44 g/mol) hence 100g of solute dissolved in 1000g 100g NaCl = 100/58.4 mol = 1.71 mol 1000g = 1000/18.015 mol = 55.51 mol. so, the mol fraction of NaCl = (1.71/55.51 + 1.71) (Molar mass of H_2O = 18.015 g.mol) = 0.0298 tildeequal 0.03 so, P_0 - P/P_0 = 0.06 (as 0.03 times 2) but NaCl is dissociated as Na^+ and Cl^- hence one molecule produce two so, i (van't Hoff function) = 2 Now P_0 - P/P_0 times 100 = decrease in pressure in percentage = 0.06 times 100 % = 6.00 % (observed value) Similarly this calculation can be done KI and Hence, the solution is 100g of NaCl dissolved in 1L solution.