$A$ solution containing $0.1050 \ g$ of a substance in $15.84 \ g$ of ether has a boiling point $1 \ ^\circ C$ higher than that of pure ether. What is the molar mass of the substance? [Given: Molar elevation constant for ether per $100 \ g = 21.6$]

  • A
    $144.5$
  • B
    $143.18$
  • C
    $140.28$
  • D
    $146.66$

Explore More

Similar Questions

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 $K$)

If an aqueous solution contains $9 \%$ and $1 \%$ $(w/w)$ of two non-volatile non-electrolytes $X$ (molecular weight $180$) and $Y$ (molecular weight $50$) respectively,the boiling point of the solution in ${ }^{\circ} C$ is approximately $(K_b = 0.52 \ K \ kg \ mol^{-1})$

The molal elevation constant of water is $0.513 \ ^oC \ kg \ mol^{-1}$. If $0.1 \ mol$ of sugar is dissolved in $200 \ g$ of water,at what temperature $(^oC)$ will the solution start to boil under $1 \ bar$ pressure?

If the boiling point of water is $100 \,^{\circ}C$,how many grams of $NaCl$ must be added to $500 \,g$ of water to increase the boiling point by approximately $1 \,^{\circ}C$ (in $,g$)? Given $(K_{b})_{H_2O} = 0.52 \,K \cdot kg/mol$.

The boiling point of water $(100\,^{\circ}C)$ becomes $100.52\,^{\circ}C$,if $3\,g$ of a non-volatile solute is dissolved in $200\,mL$ of water. The molecular weight of the solute is: (${K_b}$ for water is $0.6\,K\,kg\,mol^{-1}$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo