29) What is the minimum pressure in kPa that must be applied at 25 °C to obtain

Question

29) What is the minimum pressure in kPa that must be applied at 25 °C to obtain pure water by reverse osmosis from water that is 0.225 M in sodium chloride and 0.063 M in magnesium sulfate? Assume complete dissociation for electrolytes.

27) If an 0.880 m aqueous solution freezes at –3.50 °C, what is the van't Hoff factor, i, of the solute? Kf values can be found here.

26) Assuming 100% dissociation, calculate the freezing point and boiling point of 2.75 m Na2SO4(aq). Constants may be found here.

Explanation / Answer

29 ) M in the osmotic pressure formula comprises the total molarity
NaCl -> Na+ + Cl-
MgSO4 -> Mg+2 + SO4-2
M = 2(0.225 M) + 2(0.063 M)
M = 0.576 M
= (0.576 M)(0.08206 atm-L/mol-K)(25 + 273.15 K)
= 14.09 atm
(14.09 atm)(101.3 kPa/atm) = 1427.57 kPa

27) dTfp = kf x i x m

dTfp = normal fp - depressed fp = 0°C - (-3.50°C) = 3.50°C
Kf = cryoscopic constant for the solvent = 1.86°C/m for water
m = molality = 0.880m
i = (3.50)/(1.86 * 0.880) = 2.14 is the vant hoff factor

26) If you have 100% dissociation then a solution containing 2.75 m Na2SO4 will have 5.5 m Na+ ions and 2.75 m SO4^-2 ions. The total ionic concentration is 5.5+2.75 = 8.25

Equation: T = i*Kf*m, where T = T(solution) - T(pure solvent)
Kf = molal freezing point depression constant = 1.868°C/m; i = ions produced from salt = 3;
T(pure solvent) = 0.00°C; m = 2.92 m.

T = (3)(1.868°C/m)(2.75 m) = 15.411°C
T(solution) = T(pure solvent) - T = -15.41°C

For boiling point elevation:
Equation: T = -i*Kb*m, where T = T(solution) - T(pure solvent)
T(solution) = T(pure solvent) + T
T(pure solvent) = 100.0°C; Kb = 0.521°C/m; i = 3

T = (3)(0.521°C/m)(2.75m) = 4.30°C
T(solution) = T(pure solvent) + T = 100.00°C + 4.30°C = 104.30°C = 104.30°C

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