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The v.p. of water and diluted aq.solution are 24mmHg and 22.88mmHg .the molarity is

The v.p. of water and diluted aq.solution are 24mmHg and 22.88mmHg .the molarity is 
 

Grade:12

1 Answers

Samyak Jain
333 Points
4 years ago
Let P0 and PS denote vapour pressure of pure water and diluted aqueous solution respectively.
Let nB and nA denote number of moles of solute and solvent(water) in the solution.
P0 = 24 mm of Hg  and  PS = 22.88 mm of Hg.
XB = mole fraction of solute = nB / (nA + nB).
By Roult’s law, (P0 – PS) / P0 = nB / (nA + nB). For diluted aqueous solutions, nA + nB \approx nA
i.e. (P0 – PS) / P0 = nB / nA             [But actual formula is (P0 – PS) / PS = nB / nA]
\therefore (24 – 22.88) / 24 = nB / nA   \Rightarrow  1.12 / 24 = nB / nA  \Rightarrow  nB = 14nA / 300.
Mass of water in solution = 18 nA g , density of water = 1 g/ml = 1000 g/L.
Volume of water = mass/density = (18 nA / 1000) L = (9 nA / 500) L = Volume of the solution.
Since no. of moles of solute is very less, it would not contribute significantly in mass of the solution.
Molarity of solution = No. of moles of solute / Volume of solution in L
         =  nB / (18 nA / 1000)  =  (14nA / 300) / (9nA / 500) = (14)(5) / (9)(3)
         =  70 / 27  \approx  2.6 M.
If you use the formula (P0 – PS) / PS = nB / nA, the molarity is nearly same as above (2.7 M).

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