Diketahui:

• massa gliserin= 3 gram
• massa air = 120 gram
• K_b air = 0,52 °C/m
• $T_{\text{b}}=100,1413\ ^{\text{o}}\text{C}\ \ \ \ \ \$

Ditanyakan:

M_r gliserin?

Dijawab:

Molalitas (m), menyatakan jumlah mol zat terlarut dalam tiap kilogram pelarut.

$m=\frac{\text{mol zat terlarut}}{\text{Kg pelarut}}\ =\frac{\text{massa}}{M_{\text{r}}}\times\frac{1.000}{P\ \left(\text{gram}\text{ pelarut}\ \right)}\$

Rumus kenaikan titik didih:

$\Delta T_{\text{b}}=K_{\text{b}}\times m\$

$\Delta T_{\text{b}}=T_{\text{b}}\ -T_{\text{b}}^{\text{o}}\$

Keterangan:

$\Delta T_{\text{b}}=$ kenaikan titik didih ( ^oC)

$K_{\text{b}}=\$ tetapan kenaikan titik didih molal (°C molal^-1)

$m=$ molalitas (m)

$T_{\text{b}}\ =$ titik didih larutan (°C)

$T_{\text{b}}^{\text{o}}=\$titik didih pelarut (°C)

• Menentukan kenaikan titik didih gliserin

Titik didih air murni (T_b^o = 100 ^oC)

$\Delta T_{\text{b}}=T_{\text{b}}\ -T_{\text{b}}^{\text{o}}\$

$=100,1413\ ^{\text{o}}\text{C}\ -\ 100\ ^{\text{o}}\text{C}\ \ \$

$=0,1413\ ^{\text{o}}\text{C}\ \ \$

• Menentukan kenaikan titik didih M_r gliserin

$\Delta T_{\text{b}}=K_{\text{b}}\times m\ \$

$\Delta T_{\text{b}}=K_{\text{b}}\times\frac{\text{massa}}{M_{\text{r}}}\times\frac{1.000}{P\ \left(\text{gram}\text{ pelarut}\ \right)}\$

$0,1413\ ^{\text{o}}\text{C}=0,52\ ^{\text{o}}\text{C molal}^{-1}\times\frac{3\ \text{gram}}{M_{\text{r}}}\ \times\frac{1.000}{120\ \text{gram}}$

$0,1413\ ^{\text{o}}\text{C}=0,52\ ^{\text{o}}\text{C molal}^{-1}\times\frac{\text{25}}{M_{\text{r}}}\ \ \ \ \ \ \ \ \$

$M_{\text{r}}=\frac{0,52\ ^{\text{o}}\text{C molal}^{-1}\times25\ }{0,1413\ ^{\text{o}}\text{C}\ }\ \ \ \ \ \ \$

$M_{\text{r}}=92\ \text{gram mol}^{-1}$

Jadi massa molekul relatif gliserin berdasarkan data dalam soal adalah 92 gram/mol.