Lowering of vapour pressure is highest for

  • A
    $0.1 \ M$ Urea
  • B
    $0.1 \ M$ glucose
  • C
    $0.1 \ M \ MgSO_4$
  • D
    $0.1 \ M \ BaCl_2$

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At $300 \ K$,the vapour pressure of a solution containing $1 \ mole$ of $n-hexane$ and $3 \ moles$ of $n-heptane$ is $550 \ mm \ Hg$. At the same temperature,if one more mole of $n-heptane$ is added to this solution,the vapour pressure of the solution increases by $10 \ mm \ Hg$. What is the vapour pressure in $mm \ Hg$ of $n-heptane$ in its pure state?

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

At room temperature,a dilute solution of urea is prepared by dissolving $0.60 \ g$ of urea in $360 \ g$ of water. If the vapour pressure of pure water at this temperature is $35 \ mm \ Hg$,the lowering of vapour pressure will be: .............. $mm \ Hg$ (molar mass of urea $= 60 \ g \ mol^{-1}$)

The relative lowering of the vapour pressure is equal to the ratio between the number of

The vapour pressure of a solution of methyl alcohol and ethyl alcohol is represented as $P = 115 X_A + 140$,where $X_A$ is the mole fraction of methyl alcohol. The value of $\lim_{X_A \to 0} \frac{P_B^o}{X_B}$ is

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