Question

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

Answer 1

The freezing point of a solution made by dissolving 11.3 g of Ca(NO3)2 in 115 g of water is -3.34 °C. To find this, we calculate molality, account for the dissociation of ions and use the freezing point depression constant for water.

Step-by-step explanation:

To calculate the freezing point depression of a solution of Ca(NO3)2 in water, we first determine the molality of the solution. With the provided mass of Ca(NO3)2 (11.3 g) and its formula weight (164 g/mol), we find there are 0.0689 moles of Ca(NO3)2. Since we only have 115 g of water, to convert to kilograms, we have 0.115 kg. The molality (m) is then 0.0689 moles / 0.115 kg = 0.599 m. Since Ca(NO3)2 dissociates into three ions (Ca2+, 2NO3-), the van't Hoff factor (i) is 3.

The depression of the freezing point is determined using the formula ΔTf = i * Kf * m, where Kf is the molal freezing point depression constant for water (1.86 °C/m). So the depression is ΔTf = 3 * 1.86 °C/m * 0.599 m = 3.34 °C.

The freezing point of the solution is then 0 °C - 3.34 °C = -3.34 °C, which is the answer.

Answer 2

Answer: The freezing point of solution is -3.34°C

Step-by-step 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:

`Ca(NO_3)_2(aq.)\rightarrow Ca^(2+)(aq.)+2NO_3^-(aq.)`

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

• To calculate the molality of solution, we use the equation:

`Molality=\frac{m_(solute)* 1000}{M_(solute)* W_(solvent)\text{ in grams}}`

Where,

`m_(solute)` = Given mass of solute
`(Ca(NO_3)_2)` = 11.3 g

`M_(solute)` = Molar mass of solute
`(Ca(NO_3)_2)` = 164 g/mol

`W_(solvent)` = Mass of solvent (water) = 115 g

Putting values in above equation, we get:

`\text{Molality of }Ca(NO_3)_2=(11.3* 1000)/(164* 115)\\\\\text{Molality of }Ca(NO_3)_2=0.599m`

• To calculate the depression in freezing point, we use the equation:

`\Delta T=iK_fm`

where,

i = Vant hoff factor = 3

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

m = molality of solution = 0.599 m

Putting values in above equation, we get:

`\Delta T=3* 1.86^oC/m.g* 0.599m\\\\\Delta T=3.34^oC`

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

`\Delta T=\text{freezing point of water}-\text{freezing point of solution}`

`\Delta T` = 3.34 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

`3.34^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-3.34^oC`

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