Advertisements
Advertisements
प्रश्न
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
उत्तर
(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.
संबंधित प्रश्न
What is meant by positive deviations from Raoult's law? Give an example. What is the sign of ∆mixH for positive deviation?
What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
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%.
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?
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
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 |
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]
At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is ______.
An aqueous solution of hydrochloric acid:
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.
\[\ce{CHCl3(l) and CH2Cl2(l)}\]
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 ______.
The vapour pressure of pure liquid X and pure liquid Y at 25°C are 120 mm Hg and 160 mm Hg respectively. If equal moles of X and Y are mixed to form an ideal solution, calculate the vapour pressure of the solution.
An azeotropic mixture of two liquids will have a boiling point lower than either of the two liquids when it ______.
