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States of Matter Test - 64

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States of Matter Test - 64
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  • Question 1
    1 / -0
    Ammonium carbamate dissociates as:
    $$N{ H }_{ 2 }COON{ H }_{ 4 }(s)\quad \rightleftharpoons \quad 2N{ H }_{ 3 }(g)\quad +\quad C{ O }_{ 2 }(g)$$
    In a closed vessel containing ammonium carbamate in equilibrium, ammonia is added such that partial pressure of $$N{ H }_{ 3 }$$ now equals to the original total pressure. Calculate the ratio of partial pressure of $$C{ O }_{ 2 }$$ now to the original partial pressure of $$C{ O }_{ 2 }$$ : 
    Solution
    $$\underset {Initial}{{NH_2COONH_4}_{(s)}} \rightleftharpoons \underset {2P}{{2NH_3}_{(g)}}+\underset {P}{{CO_2}_{(g)}}$$
    $$K_P=P_{NH_3}^2 P_{CO_2}$$
    $$\Rightarrow K_P=2P.P \longrightarrow (1)$$

    Now, in second case $$P_t=3P$$
    $$\underset {Initial}{{NH_2COONH_4}_{(s)}}\rightleftharpoons \underset {3P}{{2NH_3}_{(g)}}+\underset {P'}{{CO_2}_{(g)}}$$
    $$K_P=(3P)^2(P')\longrightarrow (2)$$

    From $$(1)$$ & $$(2)$$, $$(2P)^2P=(3P)^2(P')$$
    $$\Rightarrow P'=\cfrac {4P^3}{9P^2}=\cfrac {4P}{9}$$
    $$\therefore \cfrac {P'}{P}=\cfrac {4}{9}$$
  • Question 2
    1 / -0
    At $$20^{o}C$$, the vapour pressure of $$0.1\ M$$ solution of urea is $$0.0311\ mm$$ less than that of water and the vapour pressure of $$0.1\ M$$ solution of $$KCl$$ is $$0.0574\ mm$$ less than that of water. The apparent degree of dissociation of $$KCl$$ at this dilution is :
    Solution
    The Vant Hoff factor is the property of a solute. So it is directly proportional to any colligative property of solute (like vapour pressure).

    So, vapour pressure $$\propto$$ Vant Hoff factor (i)

    $$\implies \dfrac{vapour\:pressure_{(urea)}}{vapour\:pressure_{(KCl)}}=\dfrac{i_{urea}}{i_{KCl}}$$

    But $$i_{(urea)}=1$$

    $$\implies \dfrac{vapour\:pressure_{(KCl)}}{vapour\:pressure_{(urea)}}={i_{KCl}}=\dfrac{0.0574}{0.0311}$$   ...(i)

    $$KCl\leftrightharpoons K^++Cl^-$$

       $$1$$            $$0$$        $$0$$
    $$1-\alpha$$      $$\alpha$$         $$\alpha$$

    For dissociation, we have:
    $$\alpha=\dfrac{i-1}{n-1}$$
    Here, $$n=2$$

    $$\implies \alpha=\dfrac{i-1}{2-1}$$
    $$\implies i={1+\alpha}$$

    From (i), we have:
    $$\dfrac{0.0574}{0.0311}={1+\alpha}$$

    $$\implies \alpha=0.846$$
    So, $$\%\:\alpha=\alpha\times 100=84.6\%$$
    Hence, the correct option is (B).

  • Question 3
    1 / -0
    In the given equilibrium, $${ H }_{ 2 }O(l)\rightleftharpoons { H }_{ 2 }O(g)$$ at $$100^oC$$ the vapour pressure is $$1 \ atm$$. If the volume of the container is halved, after sometime the vapour pressure becomes:
    (assuming constant temperature)
    Solution
    The vapour pressure of a liquid depends solely on the temperature and the forces of intermolecular attraction amongst the molecules. If the volume of the container is halved and the temperature is constant, then it will have no effect on the intermolecular forces and hence the vapour pressure will remain the same as before, that is , 1 atm.

    Hence, the correct answer is option D.
  • Question 4
    1 / -0
    If $${ CuSO }_{ 4 }\cdot 5H_{ 2 }O\left( s \right) \rightleftharpoons { CuSO }_{ 4 }\cdot { 3H }_{ 2 }O\ (s)+{ 2H }_{ 2 }O\ (l){ K }_{ p }=1.086\times { 10 }^{ -4 }{ atm }^{ 2 }\ at\ { 25 }^{ o }C$$ The efflorescent nature of $${CuSO}_{4}\cdot \ {5H}_{2}O$$ can be noticed when vapour pressure of $$ {H}_{2}O$$ in atmosphere is:
    Solution
    Efflorescent nature of $$CuSO_4.5H_2O$$ means the loss of some water molecules so the compound can float on the surface rather than at bottom.

    The pressure constant contains only the partial pressure of gas, the partial solid and liquid is taken unity.

    So, $$CuSO_4.5H_2O(s)\leftrightharpoons CuSO_4.3H_2O(S)+2H_2O(g)$$
    $$K_p=(P_{H_2O})^2$$
    $$(P_{H_2O})^2=1.086\times 10^{-4}atm^2$$
    $$P_{H_2O}=1.04\times 10^{-2}atm$$

    Now, $$1atm=760\:mm\:of\:Hg$$
    $$P_{H_2O}=7.92\:mm\:of\:Hg$$
    So, for the reaction to proceed forward, the $$P_{H_2O}<7.92\:mm\:of\:Hg$$
    Hence, the correct option is (B).

  • Question 5
    1 / -0
    Which of the following has least vapour presure ?
    Solution

  • Question 6
    1 / -0
    A solution of a non-volatile solute in water has a boiling point of 375.3K. Calculate the vapour pressure of water above this solution at 338K. Given, Po (water) = 0.2467 atm at 338K and Kb for water = 0.52.
  • Question 7
    1 / -0

    $$0.2\ g$$ of a gas $$X$$ occupies a volume of 0.44 liter at given pressure and temperature. Under identical conditions of $$P$$ and $$T$$, $$0.1\ g$$ of the $$CO_2$$ gas occupies $$0.32\ L$$ volume. The gas $$X$$ can be:

    Solution
    $$0.1g$$ of $$CO_2=\dfrac { 0.1 }{ M } $$ moles of $$CO_2$$ occupies $$0.32L$$.

    $$\dfrac { 0.2 }{ M } $$ moles of $$X$$ occupies $$0.44L$$

    $$PV=nRT$$

    $$V\quad \alpha \quad n$$   ----- [At constant T and P] 

    $$\dfrac { { V }_{ \left( { CO }_{ 2 } \right)  } }{ { V }_{ X } } =\dfrac { { n }_{ { CO }_{ 2 } } }{ { n }_{ X } } \Rightarrow \dfrac { 0.32 }{ 0.44 } =\dfrac { 0.1\times M }{ 44\times 0.2 } $$
                                 
                                 $$\Rightarrow M=64$$

    & Molar mass of $${ SO }_{ 2 }=32+32=64$$
  • Question 8
    1 / -0
    Which solution has the highest vapour pressure?
  • Question 9
    1 / -0
    Reaction of $${NO}$$ takes place with hydrogen if equal molar mixture of snow and hydrogen is taken at initial total pressure of 350 mm of Mercury, total pressure reduces to half its value after 121 seconds while if initial total pressure would have been 275 mm it reduces to half of the 196 seconds calculate the order of reaction?
    Solution

  • Question 10
    1 / -0

    Directions For Questions

    When a vapour consisting of a mixture of colourless gas X and a brown gas Y, at atmospheric pressure is gradually heated from $$25^oC$$, its colour is found to deepen at first and then to fade as the temperature is raised above $$160^oC$$. At $$600^oC$$, the vapour is almost colourless consisting of a gas Z and $$O_2$$ but it's colour deepens when the pressure is raised at this temperature.

    ...view full instructions

    The brown gas Y in the mixture is?
    Solution

    Given

    Colorless gas - $$X$$

    Brown gas - $$Y$$

    The bromine does not change its color at a given temperature, but nitrogen dioxide and dinitrogen tetraoxide change their color at a given temperature and give the same color as given in the question.

    Heat dinitrogen tetroxide

     $${N_2}{O_4}$$ $$\xrightarrow{{{{160}^{\rm O}}C}}$$ $$2N{O_2}$$ $$\xrightarrow{{{{600}^{\rm O}}C}}$$ $$2NO$$ $$ + $$ $${O_2}$$

    In this reaction,

    $${N_2}{O_4}$$ - dinitrogen tetroxide (colorless)

    $$2N{O_2}$$ - nitrogen dioxide (brown)

    $$2NO$$ - Nitric oxide (colorless)

    Therefore, the name of $$Y$$ gas  is dinitrogen tetroxide that is brown.

    Hence, option (C) is the correct answer.

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