Answer to Question #168811 in Organic Chemistry for Juan

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