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?
Define azeotropes.
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.
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
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.
State Raoult’s law for a solution containing volatile components. Write two characteristics of the solution which obey Raoult’s law at all concentrations.
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
On the basis of information given below mark the correct option.
(A) In bromoethane and chloroethane mixture intermolecular interactions of A–A and B–B type are nearly same as A–B type interactions.
(B) In ethanol and acetone mixture A–A or B–B type intermolecular interactions are stronger than A–B type interactions.
(C) In chloroform and acetone mixture A–A or B–B type intermolecular interactions are weaker than A–B type interactions.
On the basis of information given below mark the correct option.
On adding acetone to methanol some of the hydrogen bonds between methanol molecules break.
Using Raoult’s law explain how the total vapour pressure over the solution is related to mole fraction of components in the following solutions.
\[\ce{CHCl3(l) and CH2Cl2(l)}\]
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 correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in a molar ratio of 3 : 2 is
[At 45°C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]
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.
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.
