Calculate the molality of the solution of a nonvolatile solute if it freezes at $-0.36 \ ^{\circ}C$. [Given: $K_{f}$ for solvent $= 1.86 \ K \ kg \ mol^{-1}$]

  • A
    $0.218 \ mol \ kg^{-1}$
  • B
    $0.193 \ mol \ kg^{-1}$
  • C
    $0.401 \ mol \ kg^{-1}$
  • D
    $0.520 \ mol \ kg^{-1}$

Explore More

Similar Questions

Calculate the molality of a solution having freezing point depression $3.6 \ K$ and freezing point depression constant $4.8 \ K \ kg \ mol^{-1}$.

Calculate the depression in freezing point of a solution when $4 \,g$ of a nonvolatile solute with a molar mass of $126 \,g \,mol^{-1}$ is dissolved in $80 \,mL$ of water. $[$Cryoscopic constant of water $K_f = 1.86 \,K \,kg \,mol^{-1}]$ (in $\,K$)

$A$ solvent freezes at $17^{\circ} C$ and it has a latent heat of fusion of $180 \ J \ g^{-1}$. The molal depression constant $(K_{f})$ of the solvent is (in $K \ kg \ mol^{-1}$):

How much amount of $NaCl$ should be added to $600 \ g$ of water $(\rho=1.00 \ g / mL)$ to decrease the freezing point of water to $-0.2^{\circ} C ?$ ............. $gm$
(The freezing point depression constant for water $=2 \ K \ kg \ mol^{-1}$ )

$3 \times 10^{-3} \ kg$ acetic acid is added into $500 \ cm^{3}$ water. If dissociation of acetic acid is $23\%$ then find out depression in freezing point? $K_f$ of water $= 1.86 \ K \ kg \ mol^{-1}$ and density $= 0.997 \ g \ cm^{-3}$.

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