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

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.