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inna [77]
3 years ago
13

The compound zinc fluoride is a strong electrolyte. write the transformation that occurs when solid zinc fluoride dissolves in w

ater.
Chemistry
1 answer:
Svetach [21]3 years ago
4 0
Strong electrolytes by definition are those compounds that completely dissociates into their component ions when dissolved in water. The chemical formula for the zinc fluoride is,
               ZnF2
This means that each formula unit is composed of one atom of zinc and 2 atoms of fluoride. The ions comprising the unit are Zn²⁺ and F⁻. The dissociation is as shown below,
              ZnF₂ -->  Zn²⁺  + 2F⁻
When dissolved in water it is expected that the compound dissociates into three different ions, one Zn²⁺ and two F⁻. 
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A certain liquid X has a normal freezing point of 7.60 °C and a freezing point depression constant K= 6.90 °C-kg-mol. Calculate
Dmitry [639]

<u>Answer:</u> The freezing point of solution is -5.11°C

<u>Explanation:</u>

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

NaCl(aq.)\rightarrow Na^{+}(aq.)+Cl^-(aq.)

The total number of ions present in the solution are 2.

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

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (NaCl) = 7.57 g

M_{solute} = Molar mass of solute (NaCl) = 58.44 g/mol

W_{solvent} = Mass of solvent (liquid X) = 350.0 g

Putting values in above equation, we get:

\text{Molality of }NaCl=\frac{7.57\times 1000}{58.44\times 350.0}\\\\\text{Molality of }NaCl=0.370m

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

\Delta T=iK_fm

where,

i = Vant hoff factor = 2

K_f = molal freezing point depression constant = 6.90°C/m

m = molality of solution = 0.370 m

Putting values in above equation, we get:

\Delta T=2\times 6.90^oC/m.g\times 0.370m\\\\\Delta T=5.11^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 = 5.11 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

5.106^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-5.11^oC

Hence, the freezing point of solution is -5.11°C

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