10Cam→5Cl2g+10e 2 MnO4 (ag)+16 H'(ag)+10 e>2 Mn(ag)+8 H2Od) Add 10 Cl(ag)+2 MnO4 (ag)+16 H(ag)>5 Cl2(g)+2 Mn"(ag)+8 H2O) C5.(a)[8 marks]An aqueous solution of a detergent at a concentration of 1.00gL has an osmotic pressure of 64.8 mm Hg at 25.0C.Calculate the molecular weight of the detergent. 64.8mmHa I760mmHg/atm =0.0853atm II=MRT 0.0853atm =0.082 LatmK 'mo'(25+273r =0.00349mo1L-1 The molecular weight is therefore 0.00349mol=286.6gmor, 1.00gL (b)[8 marks]Calculate the molality of a solution that is 30.0%by weight HO The density of the solution is 1.11gmL Assume 1 L of the solution.This has a mass of 1,000 mL x 1.11 g mL=1110g The mass of H2O2 in this solution is 30.0%(1110 g)=333 g Moles of H2O2=333 g/34.0 g/mol =9.79 mol Mass of water =1110g-333 g=777g 10 Cl– (aq) → 5 Cl2(g) + 10 e– 2 MnO4 – (aq) + 16 H+ (aq) + 10 e– → 2 Mn+2 (aq) + 8 H2O(l) Add 10 Cl– (aq) + 2 MnO4 – (aq) + 16 H+ (aq) → 5 Cl2(g) + 2 Mn+2 (aq) + 8 H2O(l) C5. (a) [8 marks] An aqueous solution of a detergent at a concentration of 1.00 g L-1 has an osmotic pressure of 64.8 mm Hg at 25.0o C. Calculate the molecular weight of the detergent. 1 1 1 64.8mmHg 0.0853atm 760mmHg / atm MRT M RT 0.0853atm 0.082LatmK mol (25 273)K 0.00349molL − − − Π = = Π = Π = = + = The molecular weight is therefore 1 1 1 1.00gL 286.6gmol 0.00349molL − − − = (b) [8 marks] Calculate the molality of a solution that is 30.0% by weight H2O2(aq). The density of the solution is 1.11 g mL-1. Assume 1 L of the solution. This has a mass of 1,000 mL x 1.11 g mL-1 = 1110 g The mass of H2O2 in this solution is 30.0%(1110 g) = 333 g Moles of H2O2 = 333 g / 34.0 g/mol = 9.79 mol Mass of water = 1110 g – 333 g = 777 g