Question
A non-volatile solute 'X' completely dimerizes in water, if the temperature is below $-3.72^{\circ} \mathrm{C}$ and the solute completely dissociates as $\mathrm{X} \rightarrow \mathrm{Y}+\mathrm{Z}$ if the temperature is above $100.26^{\circ} \mathrm{C}$. In between these two temperatures (including both temperatures), the solute is neither dissociated nor associated. One mole of ' $\mathrm{X}$ ' is dissolved in $1.0 \mathrm{~kg}$ water $\left(K_{\mathrm{b}}=0.52 \mathrm{~K}-\mathrm{kg} / \mathrm{mol}, K_{\ell}=1.86 \mathrm{~K}-\mathrm{kg} / \mathrm{mol}\right)$.Identify the incorrect information related with the solution.(a) The freezing point of solution is $-1.86^{\circ} \mathrm{C}$(b) The boiling point of solution is $101.04^{\circ} \mathrm{C} .$(c) When the solution is cooled to $-7.44^{\circ} \mathrm{C}, 75 \%$ of water present initially will separate as ice.(d) When the solution is heated to $102.08^{\circ} \mathrm{C}, 50 \%$ of water present initially will escape out as vapour.
Step 1
Here, $K_f = 1.86 \, \mathrm{K \, kg/mol}$ and $m = 1 \, \mathrm{mol/kg}$, so $\Delta T_f = 1.86 \, \mathrm{K}$. The freezing point of the solution is then $0 \, \mathrm{C} - \Delta T_f = -1.86 \, \mathrm{C}$. Show more…
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A non-volatile solute 'X' completely dimerizes in water, if the temperature is below $-3.72^{\circ} \mathrm{C}$ and the solute completely dissociates as $\mathrm{X} \rightarrow \mathrm{Y}+\mathrm{Z}$ if the temperature is above $100.26^{\circ} \mathrm{C}$. In between these two temperatures (including both temperatures), the solute is neither dissociated nor associated. One mole of ' $\mathrm{X}$ ' is dissolved in $1.0 \mathrm{~kg}$ water $\left(K_{\mathrm{b}}=0.52 \mathrm{~K}-\mathrm{kg} / \mathrm{mol}, K_{\ell}=1.86 \mathrm{~K}-\mathrm{kg} / \mathrm{mol}\right)$. Identify the incorrect information related with the solution. (a) The freezing point of solution is $-1.86^{\circ} \mathrm{C}$ (b) The boiling point of solution is $101.04^{\circ} \mathrm{C} .$ (c) When the solution is cooled to $-7.44^{\circ} \mathrm{C}, 75 \%$ of water present initially will separate as ice. (d) When the solution is heated to $102.08^{\circ} \mathrm{C}, 50 \%$ of water present initially will escape out as vapour.
Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that freezes at -14.5 °C? The freezing point for pure water is 0.0 °C and Kf is equal to 1.86 °C/m. Express your answer to three significant figures and include the appropriate units. Part B If the 3.90 m solution from Part A boils at 103.45 °C, what is the actual value of the van't Hoff factor, i? The boiling point of pure water is 100.00 °C and Kb is equal to 0.512 °C/m. Express your answer numerically.
A solution of water (Kf = 1.86 °C/m) and glucose freezes at -2.35 °C. What is the molal concentration of glucose in this solution? Assume that the freezing point of pure water is 0.00 °C. Express your answer to three significant figures and include the appropriate units. m = molality Boiling points and molality Similar to the freezing-point depression, the boiling-point elevation ΔTb of a solution is quantitatively related to the molality m and the boiling-point-elevation constant Kb of the solvent by the equation ΔTb = Kb * m where the boiling-point elevation is the difference between the boiling points of the solution and the pure solvent. Part C A solution of water (Kb = 0.512 °C/m) and glucose boils at 102.56 °C. What is the molal concentration of glucose in this solution? Assume that the boiling point of pure water is 100.00 °C. Express your answer to three significant figures and include the appropriate units. m = molality
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