Self Studies

States of Matter Test - 70

Result Self Studies

States of Matter Test - 70
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    Vapour pressure of pure 'A' is 70 mm of Hg at $$25^0C$$ it forms an ideal solution is with 'B' in which mole fraction of A is 0.8. If the vapour pressure of the solution is 84 mm is Hg at $$25^0C$$, the vapour pressure of pure 'B' at $$25^0C$$ is :
  • Question 2
    1 / -0
    A solution is prepared by mixing $${\text{8}}{\text{.5}}\,{\text{g}}$$ of $${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_2}$$ and $$11.95\,{\text{g}}$$ of $${\text{CHC}}{{\text{l}}_{\text{3}}}$$. If vapour pressure of $${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$$ and $${\text{CHC}}{{\text{l}}_3}$$ at $${\text{298}}\,{\text{K}}$$ are $$415$$ and $$200\,{\text{mm}}$$ Hg respecetively, then mole fraction of $${\text{CHC}}{{\text{l}}_{\text{3}}}$$ in vapour form is:
    Solution

  • Question 3
    1 / -0

    Two liquids A and B form ideal solution.At 300K,the vapour pressure of solution containing 1 mol of A and 3 mol of B is 550mm Hg.At the same temperature,if one more mole of B is added to this solution,the vapour pressure of the solution increases by 10mm Hg.Determine the vapour pressures of A and B in their pure states.

    Solution

  • Question 4
    1 / -0
    The kinetic energy of $$1\ mole$$ of gas is equal to-
    Solution
    Kinetic enedy of gaseous molecule is calculaed using following formula.
    $$KE= \frac{3}{2}\times nRT$$.
    Above formula is given for n moles
    So for 1 mole $$n=1$$
    $$KE= \frac{3}{2}\times 1\times RT$$.
    $$KE= \frac{3}{2}\times RT$$.
    Hence option A is correct.


  • Question 5
    1 / -0
    Pressure exerted by 2 grams of helium present in a vessel is 1.5 atm. If 4 grams of gas 'X' is introduced into the same vessel keeping the condition constant, the pressure is 2.25 atm. Gas 'x' is  
    Solution

  • Question 6
    1 / -0
    The figure shows the graphs of pressure versus density for an ideal gas at two temperatures, $$T_1$$ and $$T_2$$, according to this which is correct?

  • Question 7
    1 / -0
    The value of gas constant per mole is approximately-
  • Question 8
    1 / -0
    Two gases X and Y have densities, $${d_{(x)}} = 3{d_{(y)}}$$ and molecular masses, $${M_{(x)}} = 0.5{M_{(y)}}$$. Then, the ratio of their pressures, i.e, $${P_x}:{P_y}$$ would be 
    Solution

  • Question 9
    1 / -0
    The vapour pressure of water at T(K) is 20 mm Hg. The following solutions are prepared at T(K) :
    Solution

  • Question 10
    1 / -0

    Two flasks of equal volume, connected by a narrow tube of negligible volume, contain $$2$$ moles of $$H_2$$ gas at $$1$$ atm pressure and $$300$$ K. Now, one of the flasks are heated to $$400$$ K and the other is maintained at $$300 K$$, then find final pressure: 

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now