Camphor is often used in molecular mass determination because

  • A
    It is volatile
  • B
    It is a solvent for organic substances
  • C
    It is readily available
  • D
    It has a very high cryoscopic constant

Explore More

Similar Questions

What is the cryoscopic constant of water if $5 \ g$ of glucose in $100 \ g$ of water causes a depression in freezing point of $2.15 \ K$? (Molar mass of glucose $= 180 \ g \ mol^{-1}$)

Calculate the molality of a solution of a non-volatile solute having a depression in freezing point of $0.93 \ K$ and a cryoscopic constant of the solvent of $1.86 \ K \ kg \ mol^{-1}$.

Bromoform has a normal freezing point of $7.734^{\circ} C$ and its $K_{f} = 14.4^{\circ} C / m$. $A$ solution of $2.60 \ g$ of an unknown substance in $100 \ g$ of bromoform freezes at $5.43^{\circ} C$. What is the molecular weight of the unknown substance?

Two elements $A$ and $B$ form compounds with molecular formulas $AB_2$ and $AB_4$. When $1 \ g$ of $AB_2$ is dissolved in $20 \ g$ of $C_6H_6$,the freezing point decreases by $2.3 \ K$. When $1 \ g$ of $AB_4$ is dissolved in $20 \ g$ of $C_6H_6$,the freezing point decreases by $1.3 \ K$. The molal depression constant for benzene is $5.1 \ K \ kg \ mol^{-1}$. Calculate the atomic weights of $A$ and $B$.

Difficult
View Solution

What is the freezing point depression constant of a solvent,$50 \ g$ of which contains $1 \ g$ of a non-volatile solute (molar mass $256 \ g \ mol^{-1}$) and the decrease in freezing point is $0.40 \ K$?

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