72) what is the freezing point (°c) of a solution prepared by dissolving 11.3 g of ca(no3)2 (formula weight = 164 g/mol) in 115 g of water? the molal freezing point depression constant for water is 1.86 °c/m. g

The freezing point of solution is -3.34°C

Explanation:

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of calcium nitrate follows:

The total number of ions present in the solution are 3.

Where,

= Given mass of solute = 11.3 g

= Molar mass of solute = 164  g/mol

= Mass of solvent (water) = 115 g

Putting values in above equation, we get:

where,

i = Vant hoff factor = 3

= molal freezing point depression constant = 1.86°C/m.g

m = molality of solution = 0.599 m

Putting values in above equation, we get:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

= 3.34 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

Hence, the freezing point of solution is -3.34°C

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day