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
By using Raoult’s law:
`(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
n1 = `(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
What is meant by positive deviations from Raoult's law? Give an example. What is the sign of ∆mixH for positive deviation?
Define azeotropes.
Why does a solution containing non-volatile solute have higher boiling point than the pure solvent ?
What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
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.
Calculate the mass of a non-volatile solute (molar mass 40 g mol−1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
100 g of liquid A (molar mass 140 g mol−1) was dissolved in 1000 g of liquid B (molar mass 180 g mol−1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
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 is meant by negative deviation from Raoult's law? Give an example. What is the sign of ∆mixH for negative deviation?
Some liquids on mixing form 'azeotropes'. What are 'azeotropes'?
For the reaction :
\[\ce{2NO_{(g)} ⇌ N2_{(g)} + O2_{(g)}}\];
ΔH = -heat
Kc = 2.5 × 102 at 298K
What will happen to the concentration of N2 if :
(1) Temperature is decreased to 273 K.
(2) The pressure is reduced
Define Raoult’s law for the elevation of the boiling point of a solution.
Raoult’s law states that for a solution of volatile liquids the partial pressure of each component in the solution is ____________.
Minimum boiling azeotrope is formed by the solution which showed
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.
\[\ce{CHCl3(l) and CH2Cl2(l)}\]
If the weight of the non-volatile solute urea (NH2–CO–NH2) is to be dissolved in 100 g of water, in order to decrease the vapour-pressure of water by 25%, then the weight of the solute will be ______ g.
An azeotropic mixture of two liquids will have a boiling point lower than either of the two liquids when it ______.
