$31 \ g$ of ethylene glycol $(C_2H_6O_2)$ is dissolved in $600 \ g$ of water. The freezing point depression of the solution is ($K_f$ for water is $1.86 \ K \ kg \ mol^{-1}$) (in $K$)

  • A
    $0.77$
  • B
    $1.55$
  • C
    $4.65$
  • D
    $3.10$

Explore More

Similar Questions

Calculate the molar mass of the solute when $1.5 \ g$ of a non-volatile solute is dissolved in $100 \ mL$ of a solvent having a density of $0.8 \ g \ mL^{-1}$,which lowers its freezing point by $0.75 \ K$. (Freezing point depression constant for the solvent is $5 \ K \ kg \ mol^{-1}$).

Which of the following aqueous molal solutions has the highest freezing point?

Column-$I$ (Various solutions) Column-$II$ (Freezing point)
$a$. $0.1 \, M \ BaCl_2$ solution $p$. $271 \, K$
$b$. $0.1 \, M \ NaCl$ solution $q$. $270 \, K$
$c$. $0.1 \, M \ K_3[Fe(CN)_6]$ solution $r$. $268 \, K$
$d$. $0.1 \, M \ Al_2(SO_4)_3$ solution $s$. $269 \, K$

Given: Freezing point of $0.1 \, M$ sucrose solution $= 272 \, K$ and freezing point of water $= 273 \, K$.
Which of the following options shows the correct matches?

The freezing point of a $0.01 \ m$ aqueous glucose solution is $-0.18^\circ C$. If an equal volume of $0.002 \ m$ glucose solution is added to it,the freezing point of the resulting solution will be ...... $^\circ C$.

When $4.5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water,the freezing point of the solution is lowered by $0.465^o C$. The molar mass of the solute is ....... $g/mol$. (Given $K_f = 1.86 \ K \ kg \ mol^{-1}$)

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