Question

For liquids are described in the table below. Use the second column of the table to explain the order of their freezing points and a third column to explain the order of their boiling points.

6.6 g of sucrose (C12H22O11) dissolved in 150.ml of water

6.6 g of potassium nitrate (KNO3) dissolved in 150.ml of water

6.6 g of sodium chloride (NaCl) dissolved in 150.ml of water

150.ml of pure water

Answer 1

К_Е = 0.52; K_C = 1.86;

b = $\frac{i\times n}{0.15}$ ; $\Delta T=Kb$

150.ml of pure water

T_frz = 0 ^oC; T_vap = 100 ^oC

6.6 g of sucrose (C12H22O11) dissolved in 150.ml of water

i = 1

n = 6.6 / 342.3 = 0.0193 (mol)

b = 0.0193 / 0.15 = 0.1287 (mol / kg)

$\Delta T_{vap}= 0.52\times$ 0.1287 = 0.067; T_vap = 100 + 0.067 = 100.067 ^oC

$\Delta T_{frz}=1.86\times$ 0.1287 = 0.24; T_frz = 0 - 0.24 = -0.24 ^oC

6.6 g of potassium nitrate (KNO3) dissolved in 150.ml of water

i = 2

n = 2$\times$6.6 / 101.1 = 0.13 (mol)

b = 0.065 / 0.15 = 0.867 (mol / kg)

$\Delta T_{vap}= 0.52\times$ 0.867 = 0.46; T_vap = 100 + 0.46 = 100.46 ^oC

$\Delta T_{frz}=1.86\times$ 0.4867= 1.62; T_frz = 0 - 1.62 = -1.62 ^oC

6.6 g of sodium chloride (NaCl) dissolved in 150.ml of water

i = 2

n = 2$\times$6.6 / 58.4 = 0.226 (mol)

b = 0.226 / 0.15 = 1.51 (mol / kg)

$\Delta T_{vap}= 0.52\times$ 1.51 = 0.79; T_vap = 100 + 0.79 = 100.79 ^oC

$\Delta T_{frz}=1.86\times$ 1.51= 2.81; T_frz = 0 - 2.81 = -2.81 ^oC