Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Given: Vapour pressure of pure benzene `(p_b^0)` = 50.71 mm Hg
Vapour pressure of pure toluene `(p_t^0)` = 32.06 mm Hg
Molar mass of benzene (C6H6) = 6 × 12 + 6 × 1
= 78 g mol−1
Molar mass of toluene (C6H5CH3) = 7 × 12 + 8 × 1
= 92 g mol−1
Number of moles present in 80 g of benzene = `80/78` mol
= 1.026 mol
Number of moles present in 100 g of toluene = `100/92` mol
= 1.087 mol
Mole fraction of benzene (χb) = `(1.026)/(1.026 + 1.087)`
= 0.486
Mole fraction of toluene (χ) = 1 − 0.486
= 0.514
Partial vapour pressure of benzene (pb) = `chi_b xx p_b^0`
= 0.486 × 50.71
= 24.65 mm Hg
Partial vapour pressure of toluene (pt) = `chi_t xx p_t^0`
= 0.514 × 32.06
= 16.48 mm Hg
As a result, the mole fraction of benzene in the vapour phase is as follows:
`p_b/(p_b + p_t)`
= `24.65/(24.65 + 16.48)`
= `24.65/41.13`
= 0.599
= 0.6
संबंधित प्रश्न
State Raoult’s law for the solution containing volatile components
Define azeotropes.
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.
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.
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.
What type of azeotrope is formed by positive deviation from Raoult’s law?
Some liquids on mixing form 'azeotropes'. What are 'azeotropes'?
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]
State Raoult’s law for a solution containing volatile components. Write two characteristics of the solution which obey Raoult’s law at all concentrations.
Minimum boiling azeotrope is formed by the solution which showed
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 ____________.
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]
A solution of a non-volatile solute in water freezes at −0.30°C. The vapour pressure of pure water at 298 K is 23.51 mm Hg and Kf for water is 1.86 degree/mol. The vapour pressure of rain solution at 298 K is ______ mm Hg.
An azeotropic mixture of two liquids will have a boiling point lower than either of the two liquids when it ______.
