2 , g of a non-volatile non-electrolyte solut | vedclass

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

Question

(A) The elevation in boiling point is given by the formula $\Delta T_b = K_b \cdot m$,where $K_b$ is the ebullioscopic constant and $m$ is the molalit

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The elevation in boiling point is given by the formula $\Delta T_b = K_b \cdot m$,where $K_b$ is the ebullioscopic constant and $m$ is the molality of the solution.Since the mass of the solute and the mass of the solvent are the same for both solutions,the molality $m$ is identical for both solvents $A$ and $B$.Given the ratio of ebullioscopic constants $\frac{(K_b)_A}{(K_b)_B} = \frac{1}{8}$.Therefore,the ratio of the elevation in boiling points is $\frac{(\Delta T_b)_A}{(\Delta T_b)_B} = \frac{(K_b)_A \cdot m}{(K_b)_B \cdot m} = \frac{(K_b)_A}{(K_b)_B} = \frac{1}{8}$.Comparing this to $\frac{x}{y}$,we get $x = 1$ and $y = 8$.The value of $y$ is $8$.

Similar Questions

Calculate the number of moles of nonvolatile solute dissolved in $0.3 \ kg$ solvent if $\Delta T_{b}=0.3 \ K$ and $K_{b}$ for solvent is $1.8 \ K \ kg \ mol^{-1}$.

The elevation in boiling point of a $0.25 \ molal$ aqueous solution of a substance is $(K_b = 0.52 \ K \ kg \ mol^{-1})$. (in $K$)

$13.44 \ g$ 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 $CuCl_2 = 134.4$]

Give the unit for the molal elevation constant in boiling point.

$A$ solution of non-volatile solute has a boiling point elevation of $1.75 \ K$. Calculate the molality of the solution $[K_{b} = 3.5 \ K \ kg \ mol^{-1}]$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module