Please wait...

Solutions Test - 8
Menu grid icon
Result Result point icon
Solutions Test - 8
  • Goals icon

    /

    Score
  • Trophy icon

    -

    Rank
White alarm icon Time Taken: -
Result frame illustration
  • Question 1/10
    1 / -0.25

     

    Which combination of following terms is matched correctly?
    I. Vapour pressure

    II. Intermolecular forces

    III. Latent heat of vaporisation

     

    Solutions

     

     

    (a) Liquid Vapour

    (a) If vapour pressure is high, it means equilibrium is in forward side. This is attained when intermolecular forces holding the molecules are of weak nature. Low heat is required to convert liquid into vapour. Thus, (a) is correct.

    (b) Thus, incorrect.

    (c) When vapour pressure is low, it means intermolecular forces are strong hence (c) is incorrect.

    (d) When vapour pressure is low, intermolecular forces are strong, hence large heat is required thus, (d) is incorrect,

     

     

  • Question 2/10
    1 / -0.25

     

    The solubility of urea in methanol is 17 g urea/100 mL methanol. Density of methanol is 0.792 g/mL at 293 K at which vapour pressure of methanol is 96.0 mm of Hg. Thus, vapour pressure of saturated solution of urea in methanol is  

     

    Solutions

     

     



     

     

  • Question 3/10
    1 / -0.25

     

    One or More than One Options Correct Type
    This section contains 3  multiple type questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

    Q. Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is/are  [JEE Advanced 2013]

     

    Solutions

     

     

    When binary mixture is ideal, then by Raoult ’s law

    ΔV = 0, ΔHmix = 0, no enthalpy change

    Mixing is spontaneous, thus ΔG <0 By thermodynamic property,



    Thus, on mixing, entropy increases

     

     

  • Question 4/10
    1 / -0.25

     

    Total vapour pressure of mixture of 1 mole of a volatile component A (p °A = 100 mm Hg) and 3 moles of volatile component B(p °B = 60 mm Hg) is 75 mm Hg. For such case, 

     

    Solutions

     

     

    But, actual vapour pressure = 75 mm Hg. Vapour pressure is increased. Force of attraction between A and B is smaller than between A and A and between B and B. Thus, more vaporisation due to decrease in boiling point, a case of positive deviation.

     

     

  • Question 5/10
    1 / -0.25

     

    At 35 °C, the vapour pressure of carbon disulphide (CS2 )is 512 mm Hg and that of acetone (CH3 COCH3 )is 344 mm Hg. A solution of CS2 and CH3 COCH3 in which mole fraction of CS2 is 0.25 has a total vapour pressure of 600 mm Hg. Which of the following statement(s) is/are correct?

     

    Solutions

     

     

     (A for acetone, C for carbon disulphide)

    But, actual ptotal  = 600 mm Hg

    Thus, there is positive deviation. Boiling point is lowered.
    Thus, ΔH = + ve (endothermic), i.e. heat is absorbed.

     

     

  • Question 6/10
    1 / -0.25

     

    Comprehension Type
    This section contains 2 paragraphs , each describing theory, experiments, data, etc. four questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)

    Passage I
    At 300 K, vapour pressure of pure liquid A = 575torr
    and that of pure liquid B = 390torr
    In binary mixture of liquids A and B with ideal behaviour, mole fraction of A in vapour phase =0.35
    Q. Binary mixture contains

     

    Solutions

     

     

    Let moles of A = nA and moles of B = nB

    Mole fraction of A in vapour state

    Thus, A is 27 mole %.
    Total vapour pressure of binary mixture  

     

     

  • Question 7/10
    1 / -0.25

     

    Passage I

    At 300 K, vapour pressure of pure liquid A = 575torr and that of pure liquid B = 390torr In binary mixture of liquids A and B with ideal behaviour, mole fraction of A in vapour phase =0.35

    Q. Total vapour pressure of the binary mixture is

     

    Solutions

     

     

    Let moles of A = nA and moles of B = nB

    Mole fraction of A in vapour state

    Thus, A is 27 mole %.
    Total vapour pressure of binary mixture  

     

     

  • Question 8/10
    1 / -0.25

     

    Passage II

    At 40 °C, the vapour pressure (in torr) of methanol-ethanol solution is represented by   

    where, is the mole fraction of methanol.

    Q. Difference of vapour pressures of volatile liquids (methanol-ethanol) in pure state at 40 °C is

     

    Solutions

     

     





     

     

  • Question 9/10
    1 / -0.25

     

    Passage II

    At 40 °C, the vapour pressure (in torr) of methanol-ethanol solution is represented by   

    where, is the mole fraction of methanol.

    Q.

    What is the mole fraction of methanol in vapour phase in a binary mixture with mole fraction of ethanol 0.50?

     

    Solutions

     

     

    (C)





     

     

  • Question 10/10
    1 / -0.25

     

    Matching List Type
    Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct

    100 g of binary mixture of A and B contains 60% of A. Mixture is subjected to the following operation.

    Original mixture  I    II containing vapours of A and B    containing vapours of A and B

    Match the content of Column I with their values in Column II and select the answer from the codes given below.

     

    Solutions

     

     







     

     

Close button icon
User Profile
-

Correct (-)

Wrong (-)

Skipped (-)

  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10
Mockers logo Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Arrow pointer icon
Click on Allow to receive notifications
Notification bell icon ×
Open Now