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States of Matter : Gases & Liquids Test 9
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States of Matter : Gases & Liquids Test 9
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  • Question 1/14
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    Direction (Q. Nos. 1-14) This section contains 14 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.

    Q. 50 mL of H2 gas diffuse through a small hole from a vessel in 20 min. Time taken by 40 ml. of O2 gas to diffuse under similar conditions will be

    Solutions

    Volume of hydrogen = 50 mL; Time for diffusion (t) = 20 min and volume of oxygen = 40 mL. Rare of diffusion of hydrogen (r1) = 50/20 = 2.5 mL/min Rate of diffusion of oxygen (r2) = 40/t mL/min Since the molecular mass of hydrogen (M1) = 2 and that of oxygen (M2) = 32, therefore :

  • Question 2/14
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    It takes 26 s for 10 mL of H2 gas to effuse through a porous membrane, it takes 130 s for 10 mL of an unknown gas under identical conditions to effuse. Hence, molar mass of the unknown gas (in g mol-1) is

    Solutions

    The correct answer is Option B.

  • Question 3/14
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    100 mL of O2 gas diffuse in 10 s. 100 mL of gas X diffuse in t s. Gas X and time t respectively can be

    Solutions

    t2/t1 = √M2/M1
     t1 = 10 s, M1 = 32g, t2 = t and M2 = M
    So we have t/10 = √M/32
    Putting the values of options in the above relation, we gat the correct answer. 
    On putting values of option a.
    2.5/10 = √(2/32)
    ¼ = √(1/16)
    So,LHS = RHS

  • Question 4/14
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    Rate of diffusion of LPG (mixture of n-butane and propane) is 1.25 times faster than that of SO3. Hence, mole fraction of n-butane in LPG is

    Solutions

     

    rLPG / rSO3  =  √(80/ MLPG) = 1.25

    Therefore, 80/M = (1.25)2

     Therefore, MLPG  =  51.2

    M =  (M1(butane ) X1 + M2(propane)X2)  / (X1 + X2)

    51.2 = 58X1 + 44(1-X1) / 1

    X1 ≈ 0.50

  • Question 5/14
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    Assume that the air in the space vehicle cabin consists of O2 with mole fraction, 0.20. if air contains both N2 and O2, then fraction of O2 that comes out the hole is 

  • Question 6/14
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    Consider a tube that has a frilled disk sealed in its centre so that gases can effuse through it in both directions. If the left hand side contains H2 at a pressure p and temperature T and the right hand side contains He at a pressure 2p and temperature T, then initial ratio of rate of effusion is 

  • Question 7/14
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    According to Graham’s law at a given temperature, the rates of diffusion rA / rB of gases A and B is given by (where, p and M are partial pressure and molar masses)

    [IIT JEE 1998, 2011]

    Solutions

    According to Graham's law,

    We have:

  • Question 8/14
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    If helium and methane are allowed to diffuse out of the container under the similar conditions of temperature and pressure, then the ratio of rate of diffusion of helium to methane is

    [IIT JEE 2005]

    Solutions

  • Question 9/14
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    Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. 20 dm3 of SO2 diffuses through the porous partition in 60 s. The volume of O2 (in dm3) which diffuses in 30 s will be (atomic mass of sulphur = 32 u)

    [JEE Main 2014]

  • Question 10/14
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    A vessel contains 0.5 mole each of SO2, H2 and CH4.  Its outlet was made open and closed after some time. Thus, order of partial pressure inside the vessel will be

    Solutions

    From Graham's law of diffusion, the rate of diffusion of a gas through a slit or any outlet or inlet in a container, is inversely proportional to the molecular weight of the gas.

    So , for a definite time (say t), SO2 will diffuse in the least amount through the outlet as it has highest mass among the three. Similarly the next in the increasing order will be CH4 and finally the last one is H2 will diffuse in the most amount.

    Thus, the final amount of gases in the container are in the order SO2 > CH4 > H2.

    It's clear that more the amount of the gas, more is its partial pressure.

    So, the correct order is P(SO2) > P(CH4) > P(H2)

  • Question 11/14
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    Consider the following pairs of gases A and B

     

    Q. Thus, relative rates of diffusion of gases A and B is in the order 

    Solutions

    r1 = √(44/44 = 1
    r2 = √(44/44) = 1
    r3 = √(32/48) = 0.81
    r4 = √18/20 = 0.94
    So the order is
    III < IV < I = II

  • Question 12/14
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    1 mole of helium and 3 moles of N2 exert a pressure of 16 atm. Due to a hole in the vessel in which mixture is placed, mixture leaks out. Thus, composition of the mixture effusing out initially is

    Solutions

    Here, total moles = 4+1 = 5, total pressure = 20 bar
    Partial pressure of He = 4/5×20 = 16 bar
    Partial pressure of CH4 = 1/5×20 = 4 bar
    Ratio of rate of diffusion of He and CH4 will be
    rHe/rCH4 = (pHe / pCH4)(MCH4 / MHe
    = (16/4)×(16/4)½
    = 8
    So, the rate of effusion of He is 8 times higher than CH4. Therefore Initial mixture must contain the molar ratio of He:CH4 = 8:1

  • Question 13/14
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    For 10 min each, at 27°C from two identical holes, nitrogen and an unknown gas are leaked into a common vessel of 3 L capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen gas. Thus, molar mass of the unknown gas is (g mol-1)

    Solutions

    V = 3L , T = 300 K , R = 0.0821 Lit atmJ/K mol, P = 4.18 Bar
    So, total moles of solution n =PV/RT  = 4.18 x 3 /(0.0821 x 300)  = 0.5 moles
    As moles of N2 = 0.4 
    Hence, no. of moles of unknown gas = 0.5 -0.4 = 0.1 moles
    Now, from Graham's law of diffusion,
    Rate of effusion of nitrogen (rN2) = no. of moles/ time taken = 0.4/ 10 = 0.04 mol/min
    Rate of effusion of unknown gas (runknown) =0.1/10 = 0.01 mol/min
    Since, rate of diffusion ∝ 1/√Molar mass
    rN2/runknown = √Munknown/MN2
    √(Munknown/MN2) = 4
    Munknown = 16×28 = 448 gm/mol

  • Question 14/14
    1 / -0

    At the start of an experiment, one end of a U-tube of 6 mm glass tubing is immersed in concentrated ammonia solution and the other end in concentrated hydrochloric acid. At the point in the tube where vapours of ammonia and hydrochloric acid meet, white cloud is formed. At what fraction of the distance along the tube from the ammonia solution, the white cloud first form ?

    Solutions

    d/6-d = √(36.5/17)  (d is distance from ammonia end, rate = distance/time both have same time)
    d/6-d = 1.46
    d = 3.56
    Fraction = 3.56/6  = 0.593

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