Question:
1. a solution prepared by dissolving 8.95mg of a gene fragment in 35ml of water has an osmotic pressure of 0.335 torr at 25'c . assuming that the gene fragment is a non electrolyte calculate its molar mass.
2.calculate the amount of KCl which must be added to 1 kg of water so that the frezing point is depressed by 2K. Kf for water is 1.86 Kkg mol-1.
Answer:
- The osmotic pressure is a colligative property and mathematically can be represented as
π = (nRT/V ) i,
where
π = The osmotic pressure in atmospheres
n =The number of moles of solute
R = The ideal gas constant 0.0821 L . atm/K.mol;
T = The Kelvin temperature
V = The volume of the solution
i = The van t Hoff factor.
SO,
π = (nRT/V ) i
or, n = π V/RT ( i = 1 for non electrolyte)
n = (0.335 torr) x (35 ml) x (1 atm) x (1L)/(0.821 L atm/molK) x (298.2 K) x (760 torr) x (1000 ml)
n = 6.30 x 10-7
So,
(8.95 mg) x (0.001 g/mg)/ 6.30 x 10-7 = 1.42 x 107 g/mol
2.
Depression in freezing point
ΔTf = Kf x m
Where, m = molality
m = ΔTf / Kf = 2 / 1.86 = 1.07
We know that molality of KCl = Number of moles of KCl/ solvent in Kg
So,
Number of moles of KCl required to be dissolved in 1Kg of water = 1.07 x 1 = 1.07 moles
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