Sifat koligatif adalah sifat larutan yang bergantung pada jumlah partikel zat yang terlarut, yaitu penurunan tekanan uap, kenaikan titik didih dan penurunan titik beku, dan tekanan osmosis.

$\begin{alignedat}{1}m_{\mbox{air}} & =\rho_{\mbox{air}}\times V_{\mbox{air}}\\
& =250\mbox{ mL}\\
\Delta Tf & =m\cdot kb\\
& =\frac{1,6}{180}\times\frac{1000}{250}\times1,8^{\circ}\mbox{C}\\
& =0,064^{\circ}\mbox{C}.
\end{alignedat}
$

$\begin{alignedat}{1}\Delta Tb & =Tb_{\mbox{larutan}}-Tb^{\circ}\\
& =100,936-100\\
& =0,936^{\circ}\mbox{C}\\
\Delta Tb & =m\cdot K_{b}\cdot i\\
0,056 & =\frac{\frac{1,755}{58,5}}{0,5}\times0,52\times(1+(2-1)\alpha)\\
\alpha & =0,8.
\end{alignedat}
$

Larutan yang memiliki titik didih yang paling tinggi adalah larutan yang mempunyai jumlah partikel paling banyak, yaitu dengan cara mengalikan konsentrasi dan jumlah ionnya.

$\mbox{Ca}_{3}\left(\mbox{PO}_{4}\right)_{2}\rightarrow\mbox{3Ca}^{2+}+\mbox{2PO}_{4}^{3-}$, jumlah ion = 5.

H adalah titik beku air murni, dan G adalah titik beku larutan. Jadi penurunan titik beku adalah garis G-H.

$\begin{alignedat}{1}\Delta Tb_{\mbox{glukosa}} & =m\cdot K_{b}\\
& =\frac{9}{180}\times\frac{1000}{400}\times0,52^{\circ}\mbox{C}\\
& =0,065^{\circ}\mbox{C}\\
\Delta Tb_{\mbox{glukosa+NaOH}} & =\frac{\frac{9}{180}+\left(\frac{0,2}{40}\times2\right)}{0,4}\times0,52^{\circ}\mbox{C}\\
& =0,078^{\circ}\mbox{C}\\
\Delta Tb_{\mbox{total}} & =\Delta Tb_{\mbox{glukosa+NaOH}}-\Delta Tb_{\mbox{glukosa}}\\
& =0,078-0,065^{\circ}\mbox{C}\\
& =0,013^{\circ}\mbox{C}.
\end{alignedat}
$

$\begin{alignedat}{1}\Delta Tb & =183,5-182^{\circ}\mbox{C}\\
& =1,5^{\circ}\mbox{C}\\
\Delta Tb & =m\cdot Kb\\
1,5 & =m\times3,56\\
m & =0,42\mbox{ molal}\\
\Delta Tf & =m\cdot Kf\\
& =0,42\times7,40\\
& \approx3,11^{\circ}\mbox{C}\\
Tf & =40,90-3,11\\
& =37,79^{\circ}\mbox{C}.
\end{alignedat}
$

$\begin{alignedat}{1}\Delta Tf & =0-(-0,465)\\
& =0,465^{\circ}\mbox{C}\\
\Delta Tf & =m\cdot Kf\\
\Delta Tf & =\frac{m}{M_{r}}\times\frac{1000}{m_{p}}\cdot Kf\\
0,465 & =\frac{1,5}{M_{r}}\times\frac{1000}{100}\times1,86\\
M_{r} & =60.
\end{alignedat}
$

Elektrolit kuat $\alpha=1$

$\begin{alignedat}{1}\Delta Tf & =0-(-1,116)\\
& =1,116^{\circ}C
\end{alignedat}
$

$\begin{aligned}\Delta Tf & =m\cdot K_{f}\cdot i\\
1,116 & =\frac{8,2}{164}\times\frac{1000}{250}\times1,86\times(1+(n-1)1)\\
n & =3.
\end{aligned}
$

$\begin{alignedat}{1}\Delta Tb_{A} & =m_{A}\times K_{b}\\
& =0,02K_{b}\\
\Delta Tb_{B} & =m_{B}\times K_{b}\times i\\
& =0,02\times K_{b}\times2\\
& =0,04K_{b}\\
\Delta Tb_{C} & =m_{C}\times K_{b}\\
& =0,03K_{b}\\
\Delta Tb_{D} & =m_{D}\times K_{b}\\
& =0,01\times K_{b}\times5\\
& =0,05K_{b}
\end{alignedat}
$

Urutan titik didih = kenaikan titik didihnya, sehingga : $Tb_{D}>Tb_{B}$$>Tb_{C}>Tb_{A}$$=\mbox{D}>\mbox{B}$$>\mbox{C}>\mbox{A}$

$\begin{alignedat}{1}\Delta T_{b} & =T_{b}-T_{b}^{\circ}\\
& =100,1404^{\circ}\mbox{C}-100^{\circ}\mbox{C}\\
& =0,1404^{\circ}\mbox{C}\\
\Delta T_{b} & =m\times K_{b}\times i\\
0,1404^{\circ}\mbox{C} & =\frac{2,85}{95}\times\frac{1000\mbox{ mL}}{200\mbox{ mL}}\times0,52\times i\\
i & =1,8\\
\Delta T_{f} & =m\times K_{f}\times i
\end{alignedat}
$

Penambahan 0,95 gram, sehingga

$\begin{aligned}\mbox{total massa AX}_{2} & =2,85+0,95\mbox{ gram}\\
& =3,8\mbox{ gram}
\end{aligned}
$

Pengenceran lima kali, sehingga

$\begin{aligned}\mbox{volume total larutan} & =5\times200\mbox{ mL}\\
& =1000\mbox{ mL}\\
& =1\mbox{ L}.
\end{aligned}
$

$\begin{alignedat}{1}\Delta T_{f} & =m\times K_{b}\times i\\
& =\frac{3,8}{95}\times\frac{1000\mbox{ mL}}{1000\mbox{ mL}}\times1,86\times1,8\\
& =0,1339^{\circ}\mbox{C}\\
\Delta T_{f} & =T_{f}^{\circ}-T_{f}\\
0,1339^{\circ}\mbox{C} & =0^{\circ}\mbox{C}-T_{f}\\
T_{f} & =-0,1339^{\circ}\mbox{C}.
\end{alignedat}
$

$\begin{alignedat}{1}\mbox{MgCl}_{2} & \rightarrow\mbox{Mg}^{2+}+\mbox{2Cl}^{-}\\
i & =1+(n-1)\alpha\\
& =1+(3-1)1\\
& =3\\
\Delta Tb_{1} & =100,39-100\\
& =0,39^{\circ}\mbox{C}\\
\Delta Tb_{2} & =100,624-100\\
& =0,624^{\circ}\mbox{C}\\
\Delta Tb_{1} & =\frac{m}{M_{r}}\times\frac{1000}{m_{p}}\times Kb\times i\\
0,39 & =\frac{m}{95}\times\frac{1000}{200}\times0,52\times3\\
m & =4,75\mbox{ gram}\\
\Delta Tb_{2} & =\frac{m_{tot}}{M_{r}}\times\frac{1000}{m_{p}}\times Kb\times i\\
0,624 & =\frac{4,75+m_{2}}{95}\times\frac{1000}{200}\times0,52\times3\\
m & =2,85\mbox{ gram}.
\end{alignedat}
$

$\begin{alignedat}{1}n\mbox{C}_{6}\mbox{H}_{12}\mbox{O}_{6} & =\frac{0,9}{180}\mbox{ mol}\\
& =0,005\mbox{ mol}\\
V_{\mbox{tot}} & =50+40\mbox{ mL}\\
& =90\mbox{ mL}\\
m_{\mbox{tot}} & =\rho\times V\\
& =1\times90\mbox{ gram}\\
& =90\mbox{ gram}\\
& =0,09\mbox{ kg}\\
\mbox{NaOH} & \rightarrow\mbox{Na}^{+}+\mbox{OH}^{-}\\
\Delta Tb & =100,087-100\\
& =0,087^{\circ}\mbox{C}
\end{alignedat}
$

$\begin{alignedat}{1}\Delta Tb= & \frac{\left(n_{\mbox{NaOH}}\times i\right)+n_{\mbox{C}_{6}\mbox{H}_{12}\mbox{O}_{6}}}{m_{\mbox{pelarut}}}\times Kb\\
0,087 & =\frac{\left(\frac{x}{40}\times2\right)+0,005}{0,09}\times0,52\\
x & =0,2\mbox{ gram}.
\end{alignedat}
$

$\begin{alignedat}{1}\Delta T_{f} & =2,52^{\circ}\mbox{C}\\
\Delta T_{f} & =K_{f}\cdot m\cdot i\\
2,52 & =4,8\times\frac{\frac{3,84}{40}}{m_{p}}\times(1+0,75)\\
m_{p} & =0,32\mbox{ kg}\\
& =320\mbox{ g}\\
\rho_{\mbox{alkohol}} & =\frac{m}{v}\\
& =\frac{320}{400}\mbox{ g/mL}\\
& =0,8\mbox{ g/mL}
\end{alignedat}
$

$\begin{alignedat}{1}\Delta Tb & =100,7215-100\\
& =0,7215^{\circ}\mbox{C}\\
\Delta Tb & =m\times K_{b}\times i\\
0,7215^{\circ}C & =\frac{\frac{6}{40}}{0,2}\times0,52\times i\\
i & =1,85\\
\Delta Tb_{2} & =100,177-100\\
& =0,177^{\circ}\mbox{C}
\end{alignedat}
$

$\begin{alignedat}{1}\mbox{massa NaOH} & =x\mbox{ gram}\\
\mbox{massa}\mbox{CO}\left(\mbox{NH}_{2}\right)_{2} & =(5,5-x)\mbox{ gram}\\
\Delta Tb & =m\cdot K_{b}\cdot i\\
0,177 & =\frac{\left(\frac{x}{40}\times1,85\right)+\frac{(5,5-x)}{60}}{0,4}\times0,52\\
x & =1,5\mbox{ gram}\\
\mbox{massa NaOH} & =1,5\mbox{ gram}\\
\mbox{massa}\mbox{CO}\left(\mbox{NH}_{2}\right)_{2} & =4\mbox{ gram}.
\end{alignedat}
$