$A$ solution containing $8.0 \, g$ of nicotine in $92 \, g$ of water freezes $0.925 \, ^{\circ}C$ below the normal freezing point of water. If the molal freezing point depression constant,$k_f = 1.85 \, ^{\circ}C \, kg \, mol^{-1}$,then the molar mass of nicotine is $...$
- A$16 \, g \, mol^{-1}$
- B$80 \, g \, mol^{-1}$
- C$320 \, g \, mol^{-1}$
- D$160 \, g \, mol^{-1}$
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Given that $\Delta T_f$ is the depression in freezing point of the solvent in a solution of a non-volatile solute of molality $m$,the quantity $\lim_{m \to 0} \left( \frac{\Delta T_f}{m} \right)$ is equal to:
DifficultIIT 1994
View SolutionArrange the following solutions in order of decreasing freezing points:
$(a) \ 0.075 \ M \ CuSO_4$ $(b) \ 0.060 \ M \ (NH_4)_2SO_4$
$(c) \ 0.14 \ M \ urea$ $(d) \ 0.04 \ M \ MgCl_2$
$(a) \ 0.075 \ M \ CuSO_4$ $(b) \ 0.060 \ M \ (NH_4)_2SO_4$
$(c) \ 0.14 \ M \ urea$ $(d) \ 0.04 \ M \ MgCl_2$
Easy
View SolutionGive the unit of $K_f$.
Easy
View Solution$31 \ g$ 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 $K$)
The solution containing $6 \ g$ urea (molar mass $60$) per $dm^3$ of water and another solution containing $9 \ g$ of solute $A$ per $dm^3$ water freeze at the same temperature. What is the molar mass of $A$?
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