Question

9.87 Which of the following solutions will give rise to a greater osmotic pressure at equilibrium: 5.00 g of NaCl in 350.0 mL water or 35.0 g of glucose in 400.0 mL water? For NaCl, MW = 58.5 amu; for glucose, MW = 180 amu.

Answer 1

thanks and hope you like the answer

= Vaut Osmatic pressure IT = iMRTI aut Hoff factor (no. of species present in dissolved state) M= Molarity of solution R- Gas constant = 0.0821 L.aten mot! KT To Temperature = 25°c + 273 & 298 ( Room tempecatue, assumphy for Nace , i = 2 because (Nace Nat & i ) M= mass of solute a molar mass of solute (glond) volume of solution (L) 5.009 58.5 ganol 50mL = 0.0855 mot a 0.350L = 0.244 mol/l Then I= (2) (0.144 mil) (0.0821 x. atm. mot etg (2984) = 11.95 alm For Glucose i=1 becaux (glucose is not ionu, stays as Single species Cronelako. M = 35.09 x 1 lyte / X ( - 180 g/mol M = 0.194 mot 0.4 L = 0.486 mol/L Then I = (1 (0.486 mol/L) (0.0821 L.atm. mot k) (2986) = 11.89 atm in Irace > T ghecose) Nall solution shows greater osmatic pressure than glucose solution

Answer 2

Oh hai.
I made the exercise as well and got the same answers :). But I think there's the problem that you can only have is that you can only have 3 SG in your answer, because your masses have 3 SG as well. Therefore, the results would be the same. This is what I think, but I'm not an expert. However, I think that the way in which you solved the problem is very solid!