$2.5 \ g$ of a non-volatile,non-electrolyte is dissolved in $100 \ g$ of water at $25^{\circ} C$. The solution showed a boiling point elevation by $2^{\circ} C$. Assuming the solute concentration is negligible with respect to the solvent concentration,the vapour pressure of the resulting aqueous solution is . . . . . . $mm$ of $Hg$ (nearest integer).
[Given : Molal boiling point elevation constant of water $(K_b) = 0.52 \ K \ kg \ mol^{-1}$,
$1 \ atm$ pressure $= 760 \ mm$ of $Hg$,molar mass of water $= 18 \ g \ mol^{-1}]$
[Given : Molal boiling point elevation constant of water $(K_b) = 0.52 \ K \ kg \ mol^{-1}$,
$1 \ atm$ pressure $= 760 \ mm$ of $Hg$,molar mass of water $= 18 \ g \ mol^{-1}]$
- A$702$
- B$704$
- C$705$
- D$707$
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Similar Questions
When $1.685 \ g$ of an alkali metal chloride is dissolved in $200 \ g$ water,the boiling point of the solution is measured to be $100.051 \ ^\circ C$. If the ionic solid has a crystal lattice with cation and anion radius $1.70 \ \mathring{A}$ and $1.80 \ \mathring{A}$ respectively,find the edge length of the solid assuming no defect in the crystal. Given: $K_b(H_2O) = 0.51 \ K \ kg \ mol^{-1}$,$N_A = 6 \times 10^{23}$,atomic masses: $[Li = 7, Na = 23, K = 39, Rb = 85.5, Cs = 133, Cl = 35.5]$.
Difficult
View SolutionOn dissolving $0.5 \ g$ of a non-volatile non-ionic solute to $39 \ g$ of benzene,its vapor pressure decreases from $650 \ mm \ Hg$ to $640 \ mm \ Hg$. The depression of freezing point of benzene (in $K$) upon addition of the solute is. . . . .
(Given data : Molar mass and the molal freezing point depression constant of benzene are $78 \ g \ mol^{-1}$ and $5.12 \ K \ kg \ mol^{-1}$,respectively)
(Given data : Molar mass and the molal freezing point depression constant of benzene are $78 \ g \ mol^{-1}$ and $5.12 \ K \ kg \ mol^{-1}$,respectively)
Match List-$I$ with List-$II$.
List-$I$ List-$II$ $A$. van't Hoff factor,$i$ $I$. Cryoscopic constant $B$. $k_{f}$ $II$. Isotonic solutions $C$. Solutions with same osmotic pressure $III$. $\frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$ $D$. Azeotropes $IV$. Solutions with same composition of vapour above it
Choose the correct answer from the options given below:
| List-$I$ | List-$II$ |
| $A$. van't Hoff factor,$i$ | $I$. Cryoscopic constant |
| $B$. $k_{f}$ | $II$. Isotonic solutions |
| $C$. Solutions with same osmotic pressure | $III$. $\frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$ |
| $D$. Azeotropes | $IV$. Solutions with same composition of vapour above it |
Choose the correct answer from the options given below:
The freezing point of an aqueous solution is $-0.186^oC$. If the molal elevation constant and molal depression constant of the solvent are $0.512$ and $1.86$ respectively,then the elevation in boiling point is .......... $^oC$.
Easy
View SolutionAssertion : If one component of a solution obeys Raoult's law over a certain range of composition,the other component will not obey Henry's law in that range.
Reason : Raoult's law is a special case of Henry's law.
Reason : Raoult's law is a special case of Henry's law.
MediumAIIMS 2011
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