The following data are for water: density of liquid 0.997 g cm-3 standard boilin

Question

The following data are for water:

density of liquid 0.997 g cm-3 standard boiling point 372.78 K at 1 bar vapH ° = 40.016 kJ mol!1

Assume the density and enthalpies are constant over the temperature range of interest.

(a) Calculate the temperature at which the equilibrium vapor pressure of pure liquid water is 0.835 bar. (Approximately the pressure at Reno’s elevation).

(b) How many grams of salt (NaCl) per liter of water would be required to raise the boiling point of water at 0.835 bar to 100.0°C? Assume that NaCl is completely dissociated, i.e. 2 moles of solute per mole of NaCl.

Explanation / Answer

a) Apply Clausius-Clapeyron equation: ln P2/P1 = (dHvap,m/R)(1/T1 - 1/T2 )

We get,

ln0.835/1 = (40016 J/80314 J).( 1/372.78 K   -   1/T2 K)

or, T2 = 367.64 K = 94.64 oC

b) NaCl = Na+ + Cl-

van't Hoff factor, i = no of moles after dissociation / no of moles initially present = 2/1 = 2

We know,

i = (dTb)observed / (dTb)calculated

or, 2 = (372.78-367.64)/ (dTb)calculated

(dTb)calculated = 2.68 K

(dTb)calculated = Kb.m = (RTb2. MH2O / 1000. dHvap,m)m

or, 2.68 = (8.314*367.642*18/1000*40016). m

or, molality of NaCl, m = 5.3 mole/kg solvent

For water as a solvent, 1 kg = 1 lit

So, m = 5.3 mole/lit

= 5.3 * 58.5 ....[As molar weight of NaCl = 58.5 g/mole]

= 310.05 g

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