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Solutions Test - 9
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Solutions Test - 9
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  • Question 1/10
    1 / -0.25

    The relative lowering in vapour pressure is proportional to the ratio of number of

    Solutions

  • Question 2/10
    1 / -0.25

    Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y?

    Solutions

  • Question 3/10
    1 / -0.25

    An aqueous solution of 2% non - volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

    Solutions

    Vapour pressure of pure water at boiling point (P∘ ) = 1 atm = 1.013 bar
    Vapour pressure of solution (Ps ) = 1.004 bar
    Let  Mass of solution = 100g
    Mass of solute = (w) = 2g
    Mass of solvent = 100 - 2 = 98g

  • Question 4/10
    1 / -0.25

    In the graph plotted between vapour pressure (V.P) and temperature (T).

    Solutions

    PQ corresponds to increase in vapour pressure of solvent with temperature and XY corresponds to increase in vapour pressure of solution with temperature. ΔT is the elevation in boiling point of a solution.

  • Question 5/10
    1 / -0.25

    A solution containing 12.5g of non-electrolyte substance in 185g of water shows boiling point elevation of 0.80K. Calculate the molar mass of the substance. (Kb ​= 0.52K kg mol−1 )

    Solutions

  • Question 6/10
    1 / -0.25

    If 1g of solute (molar mass = 50g mol−1 ) is dissolved in 50g of solvent and the elevation in boiling point is 1K. The molar boiling constant of the solvent is?

    Solutions

  • Question 7/10
    1 / -0.25

    2 g of sugar is added to one litre of water to give sugar solution. What is the effect of addition of sugar on the boiling point and freezing point of water?

    Solutions

    When a non-volatile solute is added to water there is elevation in boiling point and depression is freezing point.

  • Question 8/10
    1 / -0.25

    Sprinkling of salt helps in clearing the snow-covered roads in hills. The phenomenon involved in the process is

    Solutions

    When salt is spread over snow covered roads, snow starts melting from the surface because of the depression in freezing point of water and it helps in clearing the roads.

  • Question 9/10
    1 / -0.25

    Equimolar solutions in the same solvent have

    Solutions

    Equimolal solutions show same colligative properties i.e., equal elevation in boiling point and equal depression in freezing point.

  • Question 10/10
    1 / -0.25

    A 5% solution (w/W) of cane sugar (molar mass = 342 g mol-1 ) has freezing point of 271 K. What will be  the freezing point of a 5% glucose (molar mass = 18 g mol-1 ) in water if freezing point of pure water is 273.15 K?

    Solutions


    For cane sugar solution, 2.15 K = 
    For glucose solution,


    Freezing point of glucose solution = 273.15 - 4.085 = 269.07 K

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