$A$ solution of a nonvolatile solute is obtained by dissolving $1.5 \ g$ in $30 \ g$ of solvent. The boiling point elevation is $0.65 \ K$. Calculate the molal elevation constant if the molar mass of the solute is $150 \ g \ mol^{-1}$.

  • A
    $1.95 \ K \ kg \ mol^{-1}$
  • B
    $2.23 \ K \ kg \ mol^{-1}$
  • C
    $1.52 \ K \ kg \ mol^{-1}$
  • D
    $2.72 \ K \ kg \ mol^{-1}$

Explore More

Similar Questions

Calculate the amount of solute dissolved in $160 \ g$ of solvent that boils at $85^{\circ}C$. The molar mass of the solute is $120 \ g \ mol^{-1}$. $(K_{b}$ for solvent $= 2.7^{\circ}C \ kg \ mol^{-1}$ and boiling point of pure solvent $= 76^{\circ}C)$ (in $g$)

The boiling point of benzene is $353.23 \, K$. When $1.80 \, g$ of a non-volatile solute was dissolved in $90 \, g$ of benzene,the boiling point is raised to $354.11 \, K$. Calculate the molar mass of the solute. $K_{b}$ for benzene is $2.53 \, K \, kg \, mol^{-1}$.

The molal elevation constant is the ratio of the elevation in $B.P.$ to

The rise in the boiling point of a solution containing $1.8 \ g$ of glucose in $100 \ g$ of a solvent is $0.1 \ ^\circ C$. The molal elevation constant of the liquid is .......... $K/m$.

$1 \ g$ of non-volatile non-electrolyte solute is dissolved in $100 \ g$ of two different solvents $A$ and $B$ whose ebullioscopic constants are in the ratio of $1 : 5$. The ratio of the elevation in their boiling points,$\frac{\Delta T_b (A)}{\Delta T_b (B)}$ is

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