Question

Consider a 1.866M aqueous solution of tartaric acid, C4H6O6. The density of the solution is 1.1200g/mL. a. Calculate the percentage tartaric acid by mass in the solution. b. Calculate the molality of the solution. c. Calculate the mole fraction of tartaric acid in the solution.

Answer 1

M = 1.866 mol per liter

Assume a basis of V = 1 L of solution

then

mass of solution = V*D = 1.120 * 1000 = 1120 g of solution

moles in solution = 1.866 mol of C4H6O6

mass = mol*MW = 150.087*1.866 = 280.062342 g of solute

then

mass of solvent =1120 -280.062 = 839.937 g os solvent = 0.839 kg solvent

a)

% percent = mass solut e/ total mass * 100 = 280.062/1120 *100 = 25.05%

b)

molality = mol / kg solvent = 1.866/(0.839 ) = 2.224076

c)

mole frac of TA

mol frac = mol TA / total mol

mol TA = 1.866

mol solvent = 839.937 /18 = 46.663

total = 46.663+1.866 = 48.529

then

mol fract = 1.866 /48.529 = 0.038451