A solution containing 30 gm of non-volatile solute exactly in 90 gm of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 gm of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K.Calculate: (i) molar mass of the solute (ii) vapour pressure of water at 298 K.
Question
A solution containing 30 gm of non-volatile solute exactly in 90 gm of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 gm of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K.Calculate: (i) molar mass of the solute (ii) vapour pressure of water at 298 K.
Solution
(i) Molar mass of the solute:
We know that the relative lowering of vapour pressure is equal to the mole fraction of the solute.
So, ΔP/P0 = n2/n1
Where, ΔP = P0 - P (Lowering in vapour pressure) P0 = Vapour pressure of pure solvent P = Vapour pressure of the solution n2 = Number of moles of solute n1 = Number of moles of solvent
Given, P = 2.8 kPa n1 = Mass of solvent / Molar mass of solvent = 90 gm / 18 gm mol^-1 = 5 mol n2 = Mass of solute / Molar mass of solute = 30 gm / M (Let's assume the molar mass of solute is M)
Substituting these values in the equation, we get:
P0 - 2.8 kPa = (30 gm / M) / 5 mol
Let's solve this equation for M after we find the value of P0 in the next part.
(ii) Vapour pressure of water at 298 K:
After adding 18 gm of water, the total amount of water becomes 108 gm, which is 6 mol. The vapour pressure becomes 2.9 kPa.
So, we can write the equation as:
2.9 kPa = P0 (1 - (30 gm / M) / 6 mol)
Solving this equation simultaneously with the equation from the first part, we get:
P0 = 3.02 kPa and M = 60 gm mol^-1
So, the molar mass of the solute is 60 gm mol^-1 and the vapour pressure of water at 298 K is 3.02 kPa.
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