If the boiling points of two solvents X and Y | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The boiling point elevation constant $K_B$ is given by the formula: $K_B = \frac{R \cdot T_B^2 \cdot M}{1000 \cdot \Delta H_{vap}}$.Since the mole

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The boiling point elevation constant $K_B$ is given by the formula: $K_B = \frac{R \cdot T_B^2 \cdot M}{1000 \cdot \Delta H_{vap}}$.Since the molecular weights $M$ are the same and $R$ is a constant,we have $K_B \propto \frac{T_B^2}{\Delta H_{vap}}$.Given $\frac{(T_B)_X}{(T_B)_Y} = \frac{2}{1}$ and $\frac{(\Delta H)_X}{(\Delta H)_Y} = \frac{1}{2}$.Therefore,$\frac{(K_B)_X}{(K_B)_Y} = \left( \frac{(T_B)_X}{(T_B)_Y} ight)^2 	imes \frac{(\Delta H)_Y}{(\Delta H)_X}$.Substituting the values: $\frac{(K_B)_X}{(K_B)_Y} = (2)^2 	imes \left( \frac{2}{1} ight) = 4 	imes 2 = 8$.Thus,$m = 8$.

Similar Questions

Which among the following solutions has minimum boiling point elevation?

$13.44 \ g$ of $CuCl_2$ is dissolved in $1 \ kg$ of water. Determine the elevation in boiling point of the solution. [$K_b = 0.52 \ K \ kg \ mol^{-1}$,$mol \ wt$ of $CuCl_2 = 134.4$]

$A$ solution of $5.6 \ g$ nonvolatile solute in $50 \ g$ solvent has an elevation in boiling point of $1.75 \ K$. What is the molar mass of the solute $\left(K_{b} = 3 \ K \ kg \ mol^{-1}\right)$?

The 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 $........ \, g \, mol^{-1}$. $[K_b \text{ for benzene} = 2.53 \, K \, kg \, mol^{-1}]$

The latent heat of vaporisation of water is $9700 \, Cal/mole$ and if the boiling point is $100 \, ^oC$,the ebullioscopic constant of water is .......... $^oC \, kg/mol$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module