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States of Matter Test - 43

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States of Matter Test - 43
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  • Question 1
    1 / -0
    The density of $$O_{2}$$ is 16 kg/m$$^3$$ at NTP .At what temperature its density will be 14 kg/m$$^3$$ ? consider that the pressure remains constant.
    Solution
    We know that, $$PV=nRT$$
    or, $$P=\dfrac { m }{ VM } RT=\dfrac { \rho  }{ M } RT$$
    At $$NTP$$, $$T=273K$$
    $$P$$ is constant
    Therefore, $${ \rho  }_{ 1 }{ T }_{ 1 }={ \rho  }_{ 2 }{ T }_{ 2 }$$
    $$\Rightarrow 16\times 273=14\times { T }_{ 2 }$$
    $$\Rightarrow { T }_{ 2 }=312K$$ or $${ 39 }^{ 0 }C$$
    $$\therefore$$   Correct answer will be A.
  • Question 2
    1 / -0
    Two liquids, A and B, form an ideal solution. At the specified temperature, the vapour pressure of pure A is 200 mm Hg while that of pure B is 75 mm Hg. If the vapour over the mixture consists of 50 mol percent A, what is the mole percent A in the liquid?
    Solution

  • Question 3
    1 / -0
    The no. of moles per liter in the equation $$PV=nRT$$ is expressed by:
    Solution
    We know, $$PV=nRT$$
    $$\Rightarrow P= \cfrac {n}{V}RT$$
    $$\Rightarrow n/V=P/RT$$
    $$\Rightarrow \cfrac {n}{V}=\cfrac {\text{Number of moles}}{\text{Volume}}=\cfrac {P}{RT}$$
  • Question 4
    1 / -0
    The relative decrease in vapour pressure of an aqua solution containing 2 moles of $$4NaCl$$ in 3 moles of $${H}2_{O}$$ is 0.50. On reaction with $$AgNO_3$$ this solution will form:
    Solution

  • Question 5
    1 / -0
    A gas volume 100 cc. is kept in a vessel at pressure $$10^4$$ atm maintained at temperature $$24^0C$$. If now the pressure is increased to $$10^5$$ atm, keeping the temperature constant, then the volume of the gas becomes:
    Solution
    $$P_1V_1=P_2V_2$$
    $$\Rightarrow 10^4\times 100= 10^5 \times V_2$$
    $$\Rightarrow V_2=\cfrac {10^4 \times 100}{10^5}$$
    $$\Rightarrow V_2= 10 cc$$
  • Question 6
    1 / -0
    A glass tube of volume 112ml containing a gas is partially evacuated till the pressure in it drops to $$3.8\times { 10 }^{ -5 }$$ torr at 0 degree C.The number of molecules of the gas remaining in the tube?
    Solution
    volume = 0.112 L

    pressure = 3.75 atm

    temp = 273K

    gas law is PV=nRT

    $$n=\dfrac{3.75\times 0.112}{0.0821\times 273}$$

    n=0.01874 moles

    A mole has $$6.022\times 10^{23}$$ atoms or molecules of the pure substance being measured

    no. of molecules $$= 0.01874 \times 6.022\times 10^{23} =  112.5\times 10^{20}$$

    Hence, the correct option is $$\text{C}$$
  • Question 7
    1 / -0
    If $$P^o$$ and $$P^s$$ are the vapour pressure of the solvent and solution respectively. $$n_ 1$$ and $$n_2$$ are the mole fraction of the solvent and solute respectively, then?
    Solution

  • Question 8
    1 / -0
    The density of $$O_2$$ is 16 kg/m$$^3$$ at NTP. At what temperature its density will be 19? consider that the pressure remains constant.
    Solution
    As we know 
    $$PM=dRT$$
    For $$O_2$$ M is constant
    Given P is constant
    Thus $$d\alpha\cfrac{1}{T}\quad dT=constant$$
    at NTP, $$T=20°C=293K,d_1=16$$
    Given $$d_2=19,T_2?$$
    $$d_1T_1=d_2T_2\\16\times293=19\times T_2\\ T_2=246.73K\Rightarrow -26.26°C$$
  • Question 9
    1 / -0
    Vapour pressure diagram of some liquids plotted against temperature are shown below most volatile liquid is 

    Solution
    In this diagram the most volatile liquid is because is have maximum vapour pressure.

  • Question 10
    1 / -0
    An ideal gas filled at pressure of $$2 atm$$ and temp of $$300 K$$, in a balloon is kept in vacuum with in a large container wall of balloon is punctured then container temperature:
    Solution

    When the balloon is punctured,

    The gas in it expands in the large container

    Hence,

    It’s volume increases and according to Boyle’s law $$PV=cons\tan t$$. So,

    Volume increase, Pressure Decreases.

    $$\dfrac{PV}{T}$$ Is constant according to combined gas law, temperature remains constant.

    $$ w=-q $$

    $$ \Delta V=0 $$

    $$ \Delta U=q+w=0 $$

    Therefore temperature remains constant. 

    This is the required answer.
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