Advertisements
Advertisements
Question
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:
- molar mass of the solute.
- vapour pressure of water at 298 K.
Advertisements
Solution
(i) Let the molar mass of the solute be M g mol−1.
Now, the number of moles of solvent (water), n1 = `(90 g)/(18 g mol^-1)` = 5 mol
And the number of moles of solute, n2 = `(30 g)/(M mol^-1) = 30/(M mol)`
p1 = 2.8 kPa
Applying the relation,
`(p_1^0 - p_1)/p_1^0 = n_2/(n_1 + n_2)`
⇒ `(p_1^0 - 2.8)/p_1^0 = (30/M)/(5 + 30/M)`
⇒ `1 - 2.8/p_1^0 = (30/M)/((5 M + 30)/M)`
⇒ `1 - 2.8/p_1^0 = 30/(5 M + 30)`
⇒ `2.8/p_1^0 = 1 - 30/(5 M + 30)`
⇒ `2.8/p_1^0 = (5 M + 30 - 30)/(5 M + 30)`
⇒ `p_1^0/2.8 = (5 M + 30)/(5 M)` ......(i)
After the addition of 18 g of water
`n_1 = (90 + 18 g)/18` = 6 mol
p1 = 2.9 kPa
Again, applying the relation,
`(p_1^0 - p_1)/p_1^0 = n_2/(n_1 + n_2)`
⇒ `(p_1^0 - 2.9)/p_1^0 = (30/M)/((6 + 30)/M)`
⇒ `1 - 2.9/p_1^0 = (30/M)/((6M + 30)/M)`
⇒ `1 - 2.9/p_1^0 = 30/(6 M + 30) `
⇒ `2.9/p_1^0 = 1 - 30/(6 M + 30)`
⇒ `2.9/p_1^0 = (6 M + 30 - 30)/(6 M + 30)`
⇒ `2.9/p_1^0 = (6 M)/(6 M + 30)`
⇒ `p_1^0/2.9 = (6 M + 30)/(6 M)` .......(ii)
Dividing equation (i) by (ii), we have:
`2.9/2.8 = ((5 M + 30)/(5 M))/((6 M + 30)/(6 M))`
⇒ `2.9/2.8 xx (6 M + 30)/6 = (5 M + 30)/5`
⇒ 2.9 × 5 × (6m + 30) = 2.8 × 5 × (5M + 30)
⇒ 87M + 435 = 84M + 504
⇒ 3M = 69
⇒ M = 23 u
Therefore, the molar mass of the solute is 23 g mol−1.
(ii) Putting the value of ‘M’ in equation (i), we have:
`p_1^0/2.8 = (5 xx 23 + 30)/(5 xx 23)`
`=> p_1^0/2.8 = 145/115`
`=> P_1^0 = 3.53`
Hence, the vapour pressure of water at 298 K is 3.53 kPa.
RELATED QUESTIONS
In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes?
State Raoult’s law for the solution containing volatile components
Why does a solution containing non-volatile solute have higher boiling point than the pure solvent ?
The vapour pressures of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if the total vapour pressure is 600 mm Hg. Also, find the composition of the vapour phase.
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.
What type of azeotrope is formed by positive deviation from Raoult’s law?
What is meant by negative deviation from Raoult's law? Give an example. What is the sign of ∆mixH for negative deviation?
Match the following:
| (i) | Colligative property | (a) | Polysaccharide |
| (ii) | Nicol prism | (b) | Osmotic pressure |
| (iii) | Activation energy | (c) | Aldol condensation |
| (iv) | Starch | (d) | Polarimeter |
| (v) | Acetaldehyde | (e) | Arrhenius equation |
A solution containing 8.44 g of sucrose in 100 g of water has a vapour pressure 4.56 mm of Hg at 273 K. If the vapour pressure of pure water is 4.58 mm of Hg at the same temperature, calculate the molecular weight of sucrose.
What will be the vapour pressure of a solution containing 5 moles of sucrose (C12H22O11) in 1 kg of water, if the vapour pressure of pure water is 4.57 mm of Hg? [C = 12, H = 1, O = 16]
Raoult’s law states that for a solution of volatile liquids the partial pressure of each component in the solution is ____________.
The boiling point of an azeotropic mixture of water and ethanol is less than that of water and ethanol. The mixture shows ____________.
Considering the formation, breaking and strength of hydrogen bond, predict which of the following mixtures will show a positive deviation from Raoult’s law?
Using Raoult’s law explain how the total vapour pressure over the solution is related to mole fraction of components in the following solutions.
\[\ce{NaCl(s) and H2O(l)}\]
Two liquids X and Y form an ideal solution. The mixture has a vapour pressure of 400 mm at 300 K when mixed in the molar ratio of 1 : 1 and a vapour pressure of 350 mm when mixed in the molar ratio of 1 : 2 at the same temperature. The vapour pressures of the two pure liquids X and Y respectively are ______.
