An ideal solution was prepared by dissolving some amount of cane sugar (non-volatile) in 0.9 moles of water. The solution was then cooled just below its freezing temperature ( T f = 271 K ), where s

Question

An ideal solution was prepared by dissolving some amount of cane sugar (non-volatile) in 0.9 moles of water. The solution was then cooled just below its freezing temperature ( T f ​ = 271 K ), where some ice get separated out. The remaining aqueous solution registered a vapour pressure of 700 torr at 373 K . Calculate the mass (in gm) of ice separated out, if the molar heat of fusion of ice is 6 kJ / mol .

TestHub · Accepted Answer

Step 1: Calculate the molal freezing point depression constant ( $K_{f}$ ) for water using the heat of fusion:

$K_{f} = \frac{R \cdot ( T _{f}^{\circ} ) ^{2} \cdot M _{w a t er}}{1000 \cdot Δ H _{f u s}} = \frac{8.314 \times 27 3 ^{2} \times 18}{1000 \times 6000} \approx 1.86 K k g / m o l$

Step 2: Find the initial molality ( $m$ ) and initial moles of cane sugar ( $n_{s ug a r}$ ) using $Δ T_{f} = 273 - 271 = 2 K$ :

$m = \frac{Δ T _{f}}{K _{f}} = \frac{2}{1.86} = 1.075 m o l / k g$

$n_{s ug a r} = 1.075 \times (0.9 \times 0.018) \approx 0.01742 m o l$

Step 3: Use Raoult's Law ( $\frac{P ^{\circ} - P}{P} = \frac{n _{s ug a r}}{n _{w a t er, re mainin g}}$ ) at $373 K$ (where $P^{\circ} = 760 t orr$ ) to find the remaining liquid water:

$\frac{760 - 700}{700} = \frac{0.01742}{n _{w a t er, re mainin g}}$

$n_{w a t er, re mainin g} = 0.01742 \times \frac{700}{60} \approx 0.2032 m o l$

Step 4: Calculate the mass of the ice that separated out:

$M a sso f i ce = (0.9 - 0.2032) m o l \times 18 g / m o l = 12.54 g$

Chemistry - Liquid Solution Question with Solution | TestHub

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