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

The boiling point of $0.2 \ m$ aqueous urea solution is $....... \ ^oC$. ($K_b$ of water $= 0.52 \ K \ kg \ mol^{-1}$)

Which one of the following would produce maximum elevation in boiling point?

The solution containing $18 \ g \ dm^{-3}$ of glucose (molar mass $180 \ g \ mol^{-1}$) in water and another containing $6 \ g \ dm^{-3}$ of solute $A$ in water boil at the same temperature. What is the molar mass of $A$?

The order of boiling point $(B.P.)$ of the solutions formed is:

Elevation in boiling point was $0.52 \, ^\circ C$ when $6 \, g$ of a compound $X$ was dissolved in $100 \, g$ of water. The molecular weight of $X$ is: ($K_b$ for water is $0.52 \, K \, kg \, mol^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module