а a A solition is made by disolving 30g of a non- Volelite solution in 90 g of water. It has a Vapour pressure of2.8 koala at 298 k. at 298 k, Vapour pressure of pure water is 3.64 kpa. calculate the molar mass of the solution.
Answers
Answer:
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The relative lowering of vapour pressure is given by the following expression:— psa
→(p0 solvent - psolution)/ p0 solvent = n2/(n1 +n2)
Where p0 solvent is the vapour pressure of the pure solvent, psolution is the vapour pressure of a solution containing dissolved solute, n1 is the number of moles of solvent and n2 is the number of moles of solute.
For dilute solutions, n2 << n1, therefore the above expression reduces to
→ (p0 solvent - psolution)/ p0 solvent = n2/n1
= (w2 X M1) / (M2 X w1) (i)
Where w1 and w2 are the masses and M1 and M2 are the molar masses of solvent and solute respectively.
We are given that:—
w2 = 30g
w1 = 90g
psolution = 2.8kpa
p0 solvent = ? and M2 = ?
Substituting these values in relation (i), we get
(p0 solvent -2.8)/ p0 solvent = (30 X 18)/ (M2 X 90)
(p0 solvent -2.8)/ p0 solvent = 6/ M2 (1)
Similarly for second case we have the following values
w2 = 30g
w1 = (90 + 18)g = 108g
psolution = 2.9 kpa
Therefore we get,
(p0 solvent -2.9)/ p0 solvent = (30 X 18)/ (M2 X 108) = 5/M2 (2)
Dividing (1) by (2), We get
(p0 solvent -2.8)/ (p0 solvent -2.9) = 6/5
Therefore:—
p0 solvent = 3.4kpa
That is vapour pressure of water at 298K is 3.4 kpa.
Substituting the value of p0 solvent in (1), we get:—
(3.4- 2.8)/3.4 = 6/M2
or 0.6/3.4 = 6/ M2
Therefore,
M2 = 34g
Therefore,
Mass of solute is 34g and vapour pressure of water at 298K is 3.4 kpa.