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States of Matter Test - 25

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States of Matter Test - 25
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  • Question 1
    1 / -0
    'a' moles of $$PCl_5$$, undergoes, the thermal dissociation as: $$PCl_5\rightleftharpoons PCl_3+Cl_2$$, the mole fraction of $$PCl_3$$ at equilibrium is $$0.25$$ and the total pressure is $$2.0$$ atmosphere. The partial pressure of $$Cl_2$$ at equilibrium is?
    Solution
    $$PCl_5\rightarrow PCl_3+Cl_2$$
    Thus mole fraction of $$PCl_3$$ and $$Cl_2$$ are equal
    so, $$xCl_2=0.25$$
    $$P\ of\ Cl_2=x\times P_{total}=0.25\times2=0.5\ atm$$
  • Question 2
    1 / -0
    A open ended mercury manometer is used to measure the pressure exerted by a trapped gas as shown in the figure. Initially manometer shows no difference in mercury level in both columns as shown in figure.
    After sparking $$A$$ dissociates according to following reaction 
    $$A(g)\longrightarrow B(g)+3C(g)$$
    If pressure of Gas $$A$$ decreases to $$0.9atm$$. Then
    (Assume temperature to be constant and is $$300K$$)

    Solution
    Given
    Initial pressure of A gas is 1 atm
    After sparking A dissociates into B and C
    Pressure of A decreases to 0.9 atm
    Solution
    Let x be the decrease in Pressure of A
                    $$A(g)=B(g)+3C(g)$$
    t=0              1           0            0
    t=t               1-x         x            3x
    We know that
    1-x=0.9
    x= 0.1 atm
    Total pressure of system at equilibrium is
    =1 - x + x + 3x
    =1.3 atm
    The correct option is A
  • Question 3
    1 / -0
    The two bulbs of volume $$5$$ litre and $$10$$ litre containing an ideal gas at $$9\ atm$$ and $$6\ atm$$ respectively are connected. What is the final pressure in the two bulbs if the temperature remains constant ? 
    Solution

    Explanation

    Step 1: Preparing data

    • For ideal gases, the equation $$PV= nRT$$ is always valid …(1)
      where , $$P = $$ Pressure
                   $$V= $$ Volume
                   $$T =$$ Temperature
                   $$n =$$ Number of gaseous moles
                   ̣$$R =$$ Ideal gas constant
    • Let the two bulbs be A and B. The diffusion of gases will continue to take place until $$P_{1} = P_{2}$$
    • Before connecting, $$P_{1} = 9 atm$$ and $$P_{2} = 6 atm$$ … (2)
    • Also, $$V_{1} = 5 L$$ and $$V_{2} = 10 L$$ . Volumes of bulbs are fixed and do not change. …(3)

    Step 2: Solving using equations

    • Total moles of gas in the two bulbs do not change. Therefore, $$n_{total} (Initial) = n_{total} (Final)$$
      Or, $$n_{1} + n_{2} = n_{3} + n_{4}$$
    • Since the final pressure after diffusion will be equal in both bulbs, let it be $$P^{o}$$
    • Also, R and T are constant in equation (1), we can write $$n \propto PV$$
      Therefore, $$P_{1}V_{1} + P_{2}V_{2} = P^{o}V_{1} + P^{o}V_{2}$$ …(4)
    • Putting values from (2) and (3) in (4), we get:
      $$9 \times 5 + 6 \times 10 = P^{o} (V_{1} + V_{2})$$
      Or, $$45 + 60 = P^{o} (15)$$
    • Therefore, $$P^{o} = \dfrac{105}{15} = 7 $$

    Correct Option: $$B$$

  • Question 4
    1 / -0
    If the pressure of gas contained in a closed vessel is increased by 0.4% when heated by 1 degree C its initial temperature must be _____
    Solution
    $$PV=nRT$$ .....(i)

    $$P\left(1+\dfrac{0.4}{100}\right)V=nRT(T+1)$$ ....(ii)

    eqn. (ii) divided by (i)

    $$1+\dfrac{0.4}{100}=\dfrac{T+1}{T}=1+\dfrac{1}{T}$$

    $$T=\dfrac{100}{0.4}=250K$$

    Hence, the correct option is $$\text{B}$$
  • Question 5
    1 / -0
    The $$K_p$$ for $$2SO_{2(g)+O_{2(g)}}\rightleftharpoons 2SO_{3(g)}$$ is $$5.0\,atm^{-1}$$.What is the equilibrium partial pressure of $$O_2$$ if the equilibrium pressure of $$SO_2$$ and $$SO_3$$ are equal?
    Solution

    $$2SO_2 + O_2 \rightarrow2SO_3$$

    $$K_p$$ can be calculated as:

    $$K_p=\dfrac{[P_{SO_3^{2-}}]}{[P_{SO_2}^2]\times [P_{O_2}]}$$

    As given:

    $$[P_{SO_3}]=[P_{SO_2}]$$

    So,  

    $$5=\dfrac{1}{[O_2]}$$

    $$P_{O_2}=\dfrac{1}{5}$$

    $$P_{O_2}=0.2$$

    So, partial pressure of oxygen will be 0.2 atm.

    0.2 atm will be the partial pressure of $$O_2$$.

  • Question 6
    1 / -0
    The given diagram shows percentage composition of gases $$P, Q, R$$ and other gases present in air.
    Which of the following statements is/ are correct regarding gases $$P, Q$$ and $$R$$?
    (i) $$P$$ can be fixed in soil by certain bacteria present in roots of leguminous plants.
    (ii) $$P$$ supports combustion whereas $$Q$$ and $$R$$ help in extinguishing fire.
    (iii) Plants and animals release $$R$$ in respiration whereas plants use $$Q$$ in photosynthesis.
    (iv) $$R$$ entraps sun's heat and makes earth warm and hospitable.

    Solution
    According to figure P is nitrogen and nitrogen can be fixed in soil by certain bacteria present in roots of leguminous plants, and Q is oxygen and R is carbon-dioxide which entraps sun's heat and makes earth warm and hospitable.

  • Question 7
    1 / -0
    Which one of the following gases is lighter than air?
    Solution
    Hydrogen (density $$0.090$$ g/L at STP, average molecular mass $$2.016$$ g/mol) and helium (density $$0.179$$ g/L at STP, average molecular mass $$4.003$$ g/mol) are the most commonly used lift gases. Although helium is twice as heavy as (diatomic) hydrogen, they are both so much lighter than air that this difference only results in hydrogen having $$8\%$$ more buoyancy than helium.
  • Question 8
    1 / -0
    The molecular weights of $$O_2$$ and $$N_2$$ are $$32$$ and $$28$$ respectively. At $$15^0$$C, the pressure of $$1$$gm $$O_2$$ will be the same as that of $$1$$ gm $$N_2$$ in the same bottle at the temperature:
    Solution
    Temperature of $$O_{2}$$ $$=273+15=298 K$$

    Temperature of $$N_{2}$$ $$=x$$

    As $$n$$ $$\propto$$ $$\dfrac 1T$$, as pressure is constant.

    $$\Longrightarrow { n }_{ 1 }{ T }_{ 1 }={ n }_{ 2 }{ T }_{ 2 }\Longrightarrow \left( \cfrac { 1 }{ 32 }  \right) \left( 298 \right) =\left( \cfrac { 1 }{ 28 }  \right) \left( x \right) \\ \Longrightarrow x=\cfrac { 28 }{ 32 } \times 298=260.75K$$

    Temperature in $$^oC=260.75-273=-13 ^oC$$

    Hence, option $$B$$ is correct.
  • Question 9
    1 / -0
    The molecular weight of $$O_2$$ and $$N_2$$ are 32 and 28 respectively. At $$15^0C$$, the pressure of 1 gm $$O_2$$ will be the same as that of 1 gm $$N_2$$ in the same bottle at the temperature ?
    Solution
    Temperature of $$O_{2}$$ $$=273+15=298 K$$
    Temperature of $$N_{2}$$ $$=x$$
    As $$P$$$$\propto$$$$nT$$, as pressure is same.
    $$\Longrightarrow { n }_{ 1 }{ T }_{ 1 }={ n }_{ 2 }{ T }_{ 2 }\Longrightarrow \left( \cfrac { 1 }{ 32 }  \right) \left( 298 \right) =\left( \cfrac { 1 }{ 28 }  \right) \left( x \right) \\ \Longrightarrow x=\cfrac { 28 }{ 32 } \times 298=260.75K$$
    Temperature in $$^oC=260.75-273=-13 ^oC$$
  • Question 10
    1 / -0
    Lowering of vapour pressure in $$2$$ molal aqueous solution at $$373K$$ is____________.
    Solution
    Given 
    A solute B have 2 moles in 1 kg of water
    Solution
    Molality=1
    No.of moles of B=Xb
    Molality=Xb*1000/(1-Xb)*18
    Putting values
    Xb=0.0347
    No.of moles of Water=Xa
    Xa=1-Xb
    Xa=0.9653
    Pressure of 1 mole =760mmHg
    Pressure of 0.9653 mol=760*0.9653=733.628mmHg
    Lowering of Vapour pressure=760-733.628=26.372mmHg=0.035 bar
    The correct option is D
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