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 $$C = \frac{w}{M_{\ce{(NH4)2CO3}}}\cdot \frac{1}{V}$$

| improve this answer | |
  • 1
    $\begingroup$ Please do not omit the units – not only when answering homework questions; it is generally not permissible to omit the units. $\endgroup$ – user7951 Jan 20 '18 at 10:04

Not the answer you're looking for? Browse other questions tagged or ask your own question.