1.a. A saturated solution contains 54.5gCs 2 SO 4 in 30.0mL of water at 25 degre
ID: 690395 • Letter: 1
Question
1.a. A saturated solution contains 54.5gCs2SO4 in 30.0mL of water at 25 degreescelsius. What is thesolubility in g solute/(100g H2O)? Calculate the molefraction of solute at this temperature. 1.b. A saturated solution of zinc chlorate tetrahydrate,Zn(ClO3)2 *4H2O, in ethanol,C2H5OH, contains 33.4g in 20.0mL of ethanolat 25 degrees celsius. The density of ethanol at thistemperature is 0.798 g/mL. Calculate the solubility ofZn(ClO3)2 *4H2O in ethanol inunits of g solute/(100g solvent)/ 1.a. A saturated solution contains 54.5gCs2SO4 in 30.0mL of water at 25 degreescelsius. What is thesolubility in g solute/(100g H2O)? Calculate the molefraction of solute at this temperature. 1.b. A saturated solution of zinc chlorate tetrahydrate,Zn(ClO3)2 *4H2O, in ethanol,C2H5OH, contains 33.4g in 20.0mL of ethanolat 25 degrees celsius. The density of ethanol at thistemperature is 0.798 g/mL. Calculate the solubility ofZn(ClO3)2 *4H2O in ethanol inunits of g solute/(100g solvent)/ 1.b. A saturated solution of zinc chlorate tetrahydrate,Zn(ClO3)2 *4H2O, in ethanol,C2H5OH, contains 33.4g in 20.0mL of ethanolat 25 degrees celsius. The density of ethanol at thistemperature is 0.798 g/mL. Calculate the solubility ofZn(ClO3)2 *4H2O in ethanol inunits of g solute/(100g solvent)/Explanation / Answer
(a) Solubility = 54.5 g Cs2SO4 in 30.0 mL of water Water has a density ˜ 1.00 g/ mL So solubility = 54.5 g in 30.0 g of water. Solubility in 100 g water = (54.5 g Cs2SO4 ) * 100 g / 30 g = 181.67 g Cs2SO4 / 100 gH2O Moles of Cs2SO4 = 181.67 g / 361.87 g/mol = 0.502 moles Moles of water = 100 g / 18.02 g/mol = 5.55 moles Mole fraction of solute = Moles of solute / total moles = 0.502 moles / ( 0.502 + 5.55) = 0.083 . (b)33.4 g of zinc chlorate in 20.0 mL of ethanol Mass of ethanol = volume * density = 20.0 mL *0.798 g/mL = 15.96 g So solubility = 33.4 g zinc chlorate in 15.96 g ethanol Solubility in 100 g ethanol = 33.4 g zinc chlorate * (100 g / 15.96g) = 209.27 g zincchlorate / 100 g of solvent
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.