Calculate the cryoscopic constant $(K_f)$ of a solvent when $2.5 \ g$ of a solute is dissolved in $35 \ g$ of the solvent,which lowers its freezing point by $3 \ K$. (Molar mass of the solute is $117 \ g \ mol^{-1}$)
- A$3.11 \ K \ kg \ mol^{-1}$
- B$3.56 \ K \ kg \ mol^{-1}$
- C$5.52 \ K \ kg \ mol^{-1}$
- D$4.91 \ K \ kg \ mol^{-1}$
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Similar Questions
Consider the given plots of vapour pressure $(VP)$ vs temperature $(T / K)$. Which amongst the following options is the correct graphical representation showing $\Delta T_f$,the depression in the freezing point of the solvent in a solution?
What is the mass of solute having molar mass $60 \ g \ mol^{-1}$ when dissolved in $98 \ g$ of solvent decreases its freezing point by $0.2 \ K$ (in $g$)? (The numerical value of cryoscopic constant of solvent is $1.71 \ K \ kg \ mol^{-1}$)
The molar freezing point constant for water is $1.86\,^{\circ}C\,kg\,mol^{-1}$. If $342\,g$ of cane sugar $(C_{12}H_{22}O_{11})$ are dissolved in $1000\,g$ of water,the solution will freeze at $............\,^{\circ}C$.
Medium
View SolutionThe freezing point of a $4 \%$ by weight aqueous solution of $A$ is equal to the freezing point of a $10 \%$ by weight aqueous solution of $B$. If the molecular weight of $A$ is $60$,then the molecular weight of $B$ is:
Medium
View Solution$A$ solution containing $7.5 \ g$ of urea (molar mass $= 60 \ g \ mol^{-1}$) in $1 \ kg$ of water freezes at the same temperature as another solution containing $15 \ g$ of solute $X$,in the same amount of water. The molar mass of $X \ (g \ mol^{-1})$ is
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