What is the mass of (NH4)2CO3 need to add to 250 g water in order to lower the freezing point approximately 5.5 °C? [closed]








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What is the mass of $\ce{(NH4)2CO3}$ need to add to $250\ \mathrm g$ water in order to lower the freezing point approximately $5.5\ \mathrm{^\circ C}$?  

$K\_\mathrm f=1.86\ \mathrm{^\circ C\ kg/mol}$


What would be the boiling point?  

$K\_\mathrm b=0.51\ \mathrm{^\circ C\ kg/mol}$


What is the osmotic pressure (the solution in the first question) in $14.5\ \mathrm{^\circ C}$?




The van 't Hoff factor $i$ of $\ce{(NH4)2CO3}$ is $3$ in water. So $\Delta T\_\mathrm f = ik\_\mathrm fm$ can be used.


$$\Delta T\_f = 3 \cdot 1.86 \cdot \frac{w}{0.25 \cdot M\_{\ce{(NH4)2CO3}}} = \frac{3\cdot 1.86 \cdot w}{0.25\cdot 96} = 5.5$$


Solving gives $w = 23.66\ \mathrm g$.


Similarly you can use $\Delta T\_\mathrm b = ik\_\mathrm bm$ to get:


$$\Delta T\_b = 3 \cdot 0.51 \cdot \frac{w}{0.25 \cdot M\_{\ce{(NH4)2CO3}}} = \frac{\Delta T\_f \cdot 0.51}{1.86} = 1.51$$


For osmotic pressure use $\Pi = C R T$ where $C$ is molarity and is equal to approximately $$C = \frac{w}{M\_{\ce{(NH4)2CO3}}}\cdot \frac{1}{V}$$


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