Boiling point of a 2% aqueous solution of a n | vedclass

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

Question

(B) For a dilute aqueous solution,the elevation in boiling point is given by $\Delta T_B = K_B 	imes m$,where $m$ is the molality.Since the boiling p

Vedclass · Accepted Answer

$M_B = 4M_A$

Vedclass · Answer

(B) For a dilute aqueous solution,the elevation in boiling point is given by $\Delta T_B = K_B 	imes m$,where $m$ is the molality.Since the boiling points are equal,the elevation in boiling point is the same,so the molalities must be equal: $m_A = m_B$.For $A$: $2 \ g$ of solute in $98 \ g$ of solvent. Molality $m_A = \frac{2 / M_A}{98 / 1000} = \frac{2000}{98 M_A}$.For $B$: $8 \ g$ of solute in $92 \ g$ of solvent. Molality $m_B = \frac{8 / M_B}{92 / 1000} = \frac{8000}{92 M_B}$.Equating $m_A = m_B$: $\frac{2000}{98 M_A} = \frac{8000}{92 M_B}$.$\frac{1}{98 M_A} = \frac{4}{92 M_B}$.$M_B = \frac{4 	imes 98}{92} M_A = \frac{392}{92} M_A \approx 4.26 M_A$.Given the standard approximation often used in such problems where the mass of solvent is taken as $100 \ g$ (assuming dilute solution),the ratio becomes $\frac{2/M_A}{100} = \frac{8/M_B}{100}$,which leads to $M_B = 4M_A$.

Boiling point of a $2\%$ aqueous solution of a non-volatile solute $A$ is equal to the boiling point of an $8\%$ aqueous solution of a non-volatile solute $B$. The relation between molecular weights of $A$ and $B$ is:

Similar Questions

$58.4 \ g$ 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.

What is the boiling point of a solution of $0.1 \ m$ $KCl$ (in $K$)? $K_{b}$ of water is $0.52 \ K \ kg \ mol^{-1}$ and $\alpha = 100 \ \%$ (boiling point of pure water is $373 \ K$).

Calculate the boiling point elevation of a solution if $15 \ g$ of urea is dissolved in $1000 \ g$ of water. $\left[K_{b} \text{ for water} = 0.52 \ K \ kg \ mol^{-1}; \text{ molar mass of urea} = 60 \ g \ mol^{-1}\right]$ (in $K$)

Two different solutions were prepared by dissolving sucrose in water having concentrations of $1 \ m$ and $2 \ m$ respectively. The value of $(T_2 - T_1)$ will be ........... $K$. (Given: $K_b$ for water $= 0.52 \ K \ kg \ mol^{-1}$)

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