When a substance is dissolved in a solvent,the vapour pressure of the solvent decreases. It brings:
- A$A$. $A$ decrease in boiling point of the solution
- B$B$. An increase in boiling point of the solution
- C$C$. $A$ decrease in freezing point of the solution
- D$D$. An increase in freezing point of the solution
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Similar Questions
When $0.01 \ mol$ of nonvolatile solute is dissolved in a certain solvent,calculate the mass of the solvent in $kg$ if $\Delta T_{b} = 0.6 \ K$ and $K_{b}$ for the solvent $= 2 \ K \ kg \ mol^{-1}$. (in $kg$)
$58.4 \ g$ of $NaCl$ and $180 \ g$ of glucose were separately dissolved in $1000 \ mL$ of water. Identify the correct statement regarding the elevation of boiling point $(b.p.)$ of the resulting solutions.
DifficultWBJEE 2012
View SolutionThe boiling point of benzene is $353.23 \, K$. When $1.80 \, g$ of a nonvolatile solute was dissolved in $90 \, g$ of benzene,the boiling point is raised to $354.11 \, K$. The molar mass of the solute is $........ \, g \, mol^{-1}$. $[K_b \text{ for benzene} = 2.53 \, K \, kg \, mol^{-1}]$
Medium
View SolutionThe correct relation between the elevation in boiling point and the molar mass of a solute is: ($W_1$ and $W_2$ are respectively weight of solvent $\&$ solute)
$A$ solution of $6 \ g$ of solute in $100 \ g$ of water boils at $100.52^{\circ} C$. The molal elevation constant of water is $0.52 \ K \ kg \ mol^{-1}$. What is the molar mass of the solute?
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