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States of Matter Test - 24

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States of Matter Test - 24
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  • Question 1
    1 / -0
    On the surface of the earth at $$1\ atm$$ pressure, a balloon filled with $$H_{2}$$ gas occupies $$500\ mL$$. This volume is $$5/6$$ of its maximum capacity. The balloon is left in air. It starts rising. The height above which the balloon will burst if temperature of the atmosphere remains constant and the pressure decreases $$1\ mm$$ for every $$100\ cm$$ rise of height is:
    Solution
    Given:
    $$\ P_1 = 1 atm = 760 mm of Hg $$
    $$\ V_1 = 500 ml $$

    Solution:
    Let  Maximum capacity $$ \ V_2 $$= x
    According to question , 

    $$ 500 = \dfrac {5}{6} \times x $$

    $$\Rightarrow x = 500 × \dfrac{6}{5} $$

    $$\Rightarrow x = 600 $$
    From the ideal gas law , 
    $$\ P_1V_1 = P_2V_2 $$

    $$\Rightarrow 760 × 500 = \ P_2 × 600 $$

    $$\Rightarrow \ P_2 = 760 × \dfrac{500}{600} $$

    $$\Rightarrow \ P_2 = 633.33 mm of Hg $$

    Change in pressure = $$\ P_2 - \ P_1  $$

    $$\Rightarrow Change\ in\ pressure\ = 760 - 633.33 $$

    $$\Rightarrow Change\ in\ pressure\ = 126.67 mm of Hg $$

    It is given that for each 1 mm decrease,  rise = 100 cm
    Therefore,  Height =  100 × 126.67 cm
    $$\Rightarrow$$ Height = 126.67 m

    Hence, option C is correct and the answer is 126.67m.
  • Question 2
    1 / -0
    The vapour pressure of a solvent decreased by $$10\ mm$$ of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is $$0.2$$. What should be the mole fraction of the solvent, if the decrease in the vapour pressure is to be $$20\ mm$$ of mercury?
    Solution
    $$\text{Mole fraction of solute}=\cfrac { \text{lowering in vapour pressure} }{\text{ vapour pressure of solvent} } $$

    comparing under the two conditions

    $$\cfrac { 0.2 }{ \text{mole fraction of solute} } =\cfrac { 10 }{ 20 } \quad \quad $$

    or, the mole fraction of solute $$=0.4$$

    mole fraction of solvent $$=(1-0.4)=0.6$$

    Hence, the correct option is $$\text{B}$$
  • Question 3
    1 / -0
    If the mean free-path of gaseous molecule if 60 cm at a pressure of $$1\times 10^{-4}mm$$ mercury what will be its mean free-path when the pressure is increased to $$ 1\times 10^{-2}$$ mm mercury.
    Solution
    $$\begin{array}{l} \lambda =\frac { { RT } }{ { \sqrt { 2 } \pi { d^{ 2 } }NaP } }  & \lambda P=\frac { { RT } }{ { \sqrt { 2 } \pi { d^{ 2 } }Na } }  \\ { \lambda _{ 1 } }{ P_{ 1 } }={ \lambda _{ 2 } }{ P_{ 2 } } &  \\ 60\times { 10^{ -4 } }={ \lambda _{ 2 } }\times { 10^{ -2 } } &  \\ { \lambda _{ 2 } }=60\times { 10^{ -2 } }\, \, cm &  \\ { \lambda _{ 2 } }=0.6\, \, cm &  \end{array}$$
  • Question 4
    1 / -0
    An ideal gas undergoes a process in which $$T = T_{0} + aV^{3}$$, where $$T_{0}$$ and $$'a'$$ are positive constants and $$V$$ is molar volume. The molar volume for which pressure will be minimum is :
    Solution
    Given, $$T = T_0 + aV^3$$
    We know that $$PV = nRT$$
    $$\Rightarrow T =\dfrac{PV}{nR}$$
    $$\Rightarrow \dfrac{PV}{nR} = T_0 + aV^3$$
    $$\Rightarrow \dfrac{P}{nR} = \dfrac{T_0}{V} + aV^2$$
    Now, for pressure to be minimum $$\dfrac{dP}{dV} = 0$$
    $$-\dfrac{nRT_0}{aV^2} + nRa2V = 0$$
    $$V = (\dfrac{T_0}{2a})^{\dfrac{1}{3}}$$
  • Question 5
    1 / -0
    Assume that you take a flask, evacuate it to remove all the air and find it's mass to be 478.1 g .You then fill the flask with argon to a pressure of 2.15 atm and reweigh it. What would the balance read (in grams) of the flask has a volume of 7.35 L and the temperature is $$ 20^0 C$$ ?
    Solution
    Given that,
    $$P=2.15atm$$
    $$V=7.35L$$
    $$T=20.0$$
    $$R=8.3145$$ (gas constant )
    $$n=?$$
    $$PV=nRT$$
    $$2.15 \times 7.35= n\times 8.31 \times 20$$
    $$n=0.0950$$
    therefore mass =$$0.0950 \times 39.95$$
    $$3.80g$$
    But original mass =$$478.1+3.80g$$
    $$=481.90g$$
    Thus, option $$D$$ is correct. 
  • Question 6
    1 / -0
    At a constant temperature, which of the following aqueous solutions will have the maximum vapour pressure?

    [Mol weight: $$NaCl=58.5,{ H }_{ 2 }{ SO }_{ 4 }=98.0g.{ mol }^{ -1 }$$]
    Solution
    One molar (1M) aqueous solution is more concentrated than one molal aqueous solution of the same solute. 

    In solution, $${H}_{2}{SO}_{4}$$ provides three ions, while $$NaCl$$ provides two ions. Hence, the vapour pressure of a solution of $$NaCl$$ is higher (as it gives less number of ions). 

    Therefore, 1 molal $$NaCl$$ will have the maximum vapour pressure.

    Hence, option $$A$$ is correct.
  • Question 7
    1 / -0
    $$4$$g argon (Atomic mass $$=40$$) in a bulb at a temperature of TK has a pressure P atm. When the bulb was placed in hot bath at a temperature $$50^o$$C more than the first one, $$0.8$$g of gas had to be removed to get the original pressure. T is equal to :
    Solution
    We know that :

    $$PV = nRT$$

    Let us consider for argon $$P \times V = \frac{4}{{40}} \times R \times T$$

    Since,

    (molecular weight of $$Ar=40$$)

    On heating,

    $$P \times V = \frac{{3.2}}{{40}} \times R \times \left( {T + 50} \right)$$

    $$4T = 3.2T + 160$$

    $$T = \frac{{160}}{{0.8}}$$

    $$ = 200K$$

    Therefore, the correct option is B.
  • Question 8
    1 / -0
    Which of the following mixtures of gases does not obey Dalton's law of partial pressure?
    Solution
    When a mixture of $$NH_3$$ and $$HCl$$ react chemically with the spontaneous reaction it would lead in the formation of a chemical called $$NH_4Cl$$. 

    By this, the volume of the gas is automatically changed and from this reaction, there is no formation of pressure. 

    Hence the two gases do not obey Dalton's law of partial pressure.

    Hence, the correct option is $$\text{D}$$
  • Question 9
    1 / -0
    Which mixture of gases at room temp does not obey daltons law of partial pressure?
    Solution
    The mixture of $$NH_3$$ and $$HCl$$ at room temperature does not obey Dalton's law of partial pressure.

    $$NH_3$$ and $$HCl$$ react spontaneously when putting together within the same apparatus. If they react, then the volume obviously changes. Which will not result in the pressure that you would predict if the two gases don't react and remained as separate entities.

    Hence, the correct option is $$\text{C}$$
  • Question 10
    1 / -0
    A mixture of helium and neon gases is collected over water at $$28.0^o$$C and $$745$$ mmHg. If the partial pressure of helium is $$368$$ mmHg, what is the partial pressure of neon?
    [Vapour pressure of water at $$28^o C- 25.3$$ mmHg].
    Solution

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