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Boiling point of water is 373k then the ...

Boiling point of water is 373k then the boiling point of `H_2S` is

A

`gt 373k`

B

`lt 373k`

C

373k

D

`373 gt x gt 300k`

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The normal boiling point of pure water is 373 K. Calculate the boiling point of the solution containing 2.0 xx 10^(-2) kg of glucose in 0.5 kg of water ( K_(3) for water is 0.52 k kg mol^(-1) )

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Knowledge Check

  • The boiling point of water is

    A
    `0^(@)C`
    B
    `100^(@)C`
    C
    373 K
    D
    (b) and (c)

  • The boiling point of water is

    A
    `0^(@)C`
    B
    `100 ^(@)C`
    C
    `373 K`
    D
    (b) and (c )

  • If the boiling point of H2O is 373 K, the boiling point of H2S will be :

    A
    greater than 300 K but less than 373 K
    B
    equal to 373 K
    C
    more than 373 K
    D
    less than 300 K

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    A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9 Given: Freezing point depression constant of water (K_(f)^(water)=1.86 K kg mol^(-1)) Freezing point depression constant to ethanol (K_(f)^(ethanol))=2.0 K kg mol^(-1)) Boiling point elevation constant of water (K_(b)^(water))=0.52 K kg mol^(-1)) Boiling point elevation constant of ethanol (K_(b)^(ethanol))=1.2 K kg mol^(-1)) Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecular weight of water = 18 g mol^(-1) Molecular weight of ethanol = 46 g mol^(-1) In anwering the following questions consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9 . The boiling point of this solution is

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