Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर १
Number of moles of liquid A, nA = `100/140` mol
= 0.714 mol
Number of moles of liquid B, nB = `1000/180` mol
= 5.556 mol
Then, the mole fraction of A, χA = `n_A/(n_A + n_B)`
= `0.714/(0.714 + 5.556)`
= 0.114
And, mole fraction of B, χB = 1 − 0.114 = 0.886
Vapour pressure of pure liquid B, `p_B^0` = 500 torr
Therefore, the vapour pressure of liquid B in the solution,
pB = `p_B^0 chi_B`
= 500 × 0.886
= 443 torr
Total vapour pressure of the solution, ptotal = 475 torr
∴ Vapour pressure of liquid A in the solution,
pA = ptotal − pB
= 475 − 443
= 32 torr
Now,
pA = `p_A^0 chi_A`
`=> p_A^0 = p_A/chi_A`
= `32/0.114`
= 280.7 torr
Hence, the vapour pressure of pure liquid A is 280.7 torr.
उत्तर २
Number of moles of liquid A = `100/140`
= 0.714
Number of moles of liquid B = `1000/180`
= 5.555
Hence, mole fraction of A in the solution
`chi_A = 0.714/(0.714 + 5.555)`
= 0.114
and mole fraction of B in the solution
`chi_B = 5.555/(0.714 + 5.555)`
= 0.886
According to Raoult’s law,
`p_A = chi_Ap_A^circ`
`p_A = p_A^circ` ...(i)
`p_A = chi_Bp_B^circ`
= 0.886 × 500
= 443 torr ...(ii)
∴ Total vapour pressure of the solution
p = pA + pB
= `0.114 p_A^circ + 443`
Given that p = 475 torr
∴ `0.114 p_A^circ + 443 = 475`
or `p_A^circ = (475 - 443)/0.114`
= 280.7 torr
Putting this value in eq. (i), we have
pA = 0.114 × 280.7
= 31.9998
= 32 torr
संबंधित प्रश्न
In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes?
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.
Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
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?
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 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]
The boiling point of an azeotropic mixture of water and ethanol is less than that of water and ethanol. The mixture shows ____________.
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.
\[\ce{CHCl3(l) and CH2Cl2(l)}\]
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 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]
