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Vapour pressure of pure water at 298 k i...

Vapour pressure of pure water at 298 k is 23.8 mm Hg. Calculate the lowering of vapour pressure caused by adding 5g of sucrose in 50 g of water.

Text Solution

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The correct Answer is:
0.124 mm Hg
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Knowledge Check

  • The vapour pressure of water at room temperature is 23.8 mm Hg. The vapour pressure of an aqueous solution of sucrose with mole fraction 0.1 is equal to :

    A
    23.9 mm Hg
    B
    24.2 mm Hg
    C
    21.42 mm Hg
    D
    31.44 mm Hg

  • A Solution of a non-volatile solute in water freezes at -0.30^@C . The vapour-pressure of pure water at 298K is 23.51 mm Hg and K_f for water is 1.86 K. kg "mol"^(-1) Calculate the vapour-pressure of this solution at 298K.

    A
    23.4 mm
    B
    24.8 mm
    C
    34.8 mm
    D
    40 mm

  • The vapour pressure of pure water at 25^@C is 30 mm. The vapour pressure of 10% (W/W) glucose solution at 25^@C is

    A
    31.5 mm
    B
    30.6 mm
    C
    29.67 mm
    D
    26.56 mm

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    Vapour pressure of of water at 293K is 17.535 mm Hg. Calculate the vapour pressure of the solution at 293K when 25g of glucose is dissolved in 450g of water ?

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    Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50g urea (NH_(2)CONH_(2)) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering. Consider Raoult's law and formula for relative lowering in vapour pressure, (P_(A)^(0)-P_(s))/(P_(A)^(0))=(n_(B))/(n_(A))=(W_(B))/(M_(B))xx(M_(A))/(W_(A)) Where, (P_(A)^(0)-P_(s))/(P_(A)^(0)) is called relative lowering in vapour pressure.

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    The vapour pressure of pure water at 25^(@)C is 30 mm. The vapour pressure of 10% (W/W) glucose solution at 25^(@)C is

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