English
Karnataka Board PUCPUC Science 2nd PUC Class 12

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

Advertisements
Advertisements

Question

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.

Numerical
Advertisements

Solution

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 mol1

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 1: Solutions - 'NCERT TEXT-BOOK' Exercises [Page 129]

APPEARS IN

Nootan Chemistry [English] Class 12 ISC
Chapter 1 Solutions
'NCERT TEXT-BOOK' Exercises | Q 2.38 | Page 129
NCERT Chemistry Part 1 and 2 [English] Class 12
Chapter 1 Solutions
Exercises | Q 1.38 | Page 30

RELATED QUESTIONS

In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes?


State Raoult’s law for the solution containing volatile components


Why does a solution containing non-volatile solute have higher boiling point than the pure solvent ?


The vapour pressures of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if the total vapour pressure is 600 mm Hg. Also, find the composition of the vapour phase.


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%.


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:

  1. molar mass of the solute.
  2. vapour pressure of water at 298 K.

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?


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

At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is ______.


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.

\[\ce{CHCl3(l) and CH2Cl2(l)}\]


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.


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×