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Transcript
114
π 1V 1 = n1RT1
If π1 = π2
and
T1 = T2,
π2V2 = n2RT2
(Isotonic Solutions)
n1
n2
Then , V = V
1
2
Hence, when osmotic pressure and temperature are same, equal volumes of solutions would
contain equal number of moles of solute.
*9.
Ans.
Derive van’t Hoff equation for osmotic pressure of a solution.
Vant-Hoff’s equation for dilute solutions :
According to Van’t Hoff-Boyle’s law
1
π∝
(at constant temperature)
V
According to Van’t Hoff-Charle’s law,
π ∝ T
(at given concentration)
Combining the above results, we get
T
π∝
V
or πV = kT
…(i)
where k is called general solution constant.
Vant Hoff further proved that the value of k is same as R,
Hence, πV = RT
Further if the solution contains n moles of the solute dissolved in V litres of solution, then
n
C =
V
Substituting the value of C in equation (1), we get;
n
T
V
πV = n RT
π
*10.
Ans.
∴
= R
...(ii)
How is molar mass of a solute determined from osmotic pressure measurements.
According to van’t Hoff equation,
π = CRT
n
But C =
V
where n is the number of moles of solute dissolved in V litre of the solution.
n
RT
π=
V
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I