At 22 degree C the osmotic pressure of an aqueous sucrose (C_12H_22O_11) solutio

Question

At 22 degree C the osmotic pressure of an aqueous sucrose (C_12H_22O_11) solution is 3628 mm of Hg. Calculate the freezing point of the solution. The density of the solution is 1.16 g/ml. and K_f for water is 1.86 degree C.m^-1 If a voltaic (galvanic) cell is constructed using the following half - reactions Cu^2+ + e^- rightarrow Cu^+(s) E^0 = +0.15V Pb^2+ + 2e^- rightarrow Pb(s) E^0 = -0.13V Calculate the equilibrium constant for the following reaction at 25 degree C 1/2Cu^+(s) + 1/4Pb^2+ rightarrow 1/2 Cu^2+ + 1/4Pb(s) The osmotic pressure exerted by a weak acid, HA, is 2.5 atm at 25 degree C. Given the delta G degree for the dissociation of the acid as 35kJ. calculate the pH of the acid. An aqueous solution of a monoprotic acid has a pH of 2.85. Given the following values, calculate the boiling point of the solution. K_a of the acid = 1.65 times 10^-6 Density of the solution = 0.993g/ml. Molar Mass of the acid = 74.0 g/mole K_b for water = 0.512 degree C/m

Explanation / Answer

Q1.

Posm = 3628 mm Hg

find freezing point of solution

P = M*R*T

M = P/(RT) = 3628//(62.3*)22+273)) = 0.197404 mol per liter

assume a basis of

1 L = 1000 mL so

mass of solution = D*V = 1000*91.16 = 1160 g

0.197404 mol of sucrose --> mass = mol*MW = 0.197404*342.2965 = 67.570 g of sucrose

mass of water = 1160 -67.570 = 1092.43 g of water = 1.092 kg

so

molality = mol of S / kg solvent = 0.197404/1.092 = 0.180772 molal

dTf = -KF*m

dTf = -1.86*0.180772

dTF = --0.3362°C

Tf = -0.3362 °C

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