Solubility and Solutions Report Sheet PART I.Effect of Temperature on Solubility
ID: 575233 • Letter: S
Question
Solubility and Solutions Report Sheet PART I.Effect of Temperature on Solubility Water +0.9 g NaCI Water+0.9g NaCi Water+0.5 g NaCI (Room temp) (Room temp)(Hoalegy 23 4 Temperature22.1 Observations Salt did not 35.2 well w/ water discoineek inte the Solubility of NaCl at 35 2 'cis o. 0.9 -g PART I: Solution Concentration Units Mass of NaCl in Solution 1: 0.12 g Mass of H20 in Solution 1: 100g Moles of NaCl in Solution 1: mol Solution 1 Solution 2 Solution 3 Molarity Molality Mass Percent Mole Fraction Calculations:Explanation / Answer
Part II:
Mass of NaCl in solution 1 = 0.62 g
Mass of H2O in solution 1 = 100 g.
Molar mass of NaCl = (1*22.9897 + 1*35.453) g/mol = 58.4427 g/mol.
Moles of NaCl in solution 1 = (mass of NaCl in solution 1)/(molar mass of NaCl) = (0.62 g)/(58.4427 g/mol) = 0.01061 mole (ans).
Assume density of water = 1 g/mL.
Mass of the solution = mass of NaCl + mass of water = 0.62 g + 100 g = 100.62 g.
Assume that NaCl doesn’t alter the volume of the solution much; hence, volume of the solution = (mass of solution)/(density of water) = (100.62 g)/(1 g/mL) = 100.62 mL = (100.62 mL)*(1 L/1000 mL) = 0.10062 L.
Molarity of NaCl solution = (moles of NaCl)/(volume of solution in L) = (0.01061 mole)/(0.10062 L) = 0.1054 mol/L = 0.1054 M (ans).
Molality of NaCl solution = (moles of NaCl)/(kg of solvent) = (0.01061 mole)/[(100 g)*(1 kg/1000 g)] = 1.061m (ans).
Mass percent NaCl in the solution = (mass of NaCl)/(mass of solution)*100% = (0.62 g)/(100.62 g)*100% = 0.6162% (ans).
Molar mass of H2O = (2*1.008 + 1*15.9994) g/mol = 18.0154 g/mol.
Mole(s) of H2O in 100 g H2O = (100 g)/(18.0154 g/mol) = 5.5508 mole.
Mole fraction NaCl in the solution = (moles of NaCl)/(total number of moles) = (0.01061 mole)/[(0.01061 + 5.5508) mole] = 0.001907 0.0191 (ans).
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.