Question

3. A student carried out the experiment Ways to Express Concentration and obtained the following data expressed in the table. In it, he prepared a solution of NaCl (58.44 g/mol) in water (18.01 g/mol). He then extracted a 2.00 mL sample of this solution, transferred it to a test tube, and evaporated all of the solvent. With the data in the table, answer the following questions: Data: Used solute and NaCl molar mass (58.44 g/mol) Empty tube mass (E) 47.85 Tube mass with solution (g) 49.95 Tube mass with residue (6) 48.53 Volume of used solution (ML) 2.00 Volume of used solution (L)? Residual mass ()? Solution masse? Solvent mass (g)? Males solute mall? Moles solvent (mail? a) The molarity of the solution, calculated using the data from the experiment, is: a. 0.0202 b. 6.0 c.2.02 d. 3.25 b) The mole fraction of the solute is: a. 0.00650 b. 0.909 C. 0.13 d. 0.0909 c) The mole fraction of the solvent is: a. 0.550 b. 0.87 c. 0.128 d. 1.01 d) The percent by mass of solute is: 0.38 96 b 12.396 d. 32 %

Answer 1

= 12.1 g Mass of solution = 49.95 g - 47.85 g volume of solution = 2.00 ml so, density of solution (d) = mass Volume = 2.1g 2.00 ml = 1.05 g/mL • Residue = Nace So, mass of Nace in 2.1 g solution = 48.53 g -47.85 g = 10.68 g = mass of solvent C H20) = 2.1 g -0.68 g = 1.429 - = of Moles solute, Nace 0.68 g 58.44 glmol =10.0116 moll . Moles n solvent, H₂O = 1.42 g - 18.01 glmol = 0.0788 moll a) Molarity of solution = moles Vol of of solute solution in L. = 5.8 M 0.0116 mol - 2.00 mL 1000 mL/L = 16.0 ML

b) Mole fraction of solute moles of solute moles of solute & moles of solvent 0.0116 mol (0.0116 +0.0788) mol = 0.128 D: 128 = 0.13 0:13 c) mole fraction of solvent .= 0.0788 mol (0.0788 +0.0116) mol Ci- mole fraction of solute) d') % mass of solute = - X 100 mass of Mass of solute solution - 0.68 g 1 211g +100 = 32.4% =