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Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mLmL of water to produce a solution that freezes at −−14.5 ∘C? The freezing point for pure water is 0.0 ∘C∘C and Kf is equal to 1.86 ∘C/m.If the 3.90 m solution from Part A boils at 103.45 ∘C, what is the actual value of the van't Hoff factor, i? The boiling point of pure water is 100.00 ∘C and Kb is equal to 0.512 ∘C/m.

Respuesta :

Answer:

1) 108.27 grams of potassium nitarte must be added to 275 mL of water to produce a solution that freezes at -14.5°C.

2) 1.73 is the actual value of the van't Hoff factor, i.

Explanation:

1) Formula used depression in freezing point ;

[tex]\Delta T_f=T-T_f[/tex]

[tex]\Delta T_f=i\times K_f\times m[/tex]

where,

[tex]T_f[/tex] =Freezing point of solution

T = Freezing point of water

[tex]\Delta T_f[/tex] =depression in freezing point =  

i = van't Hoff factor of solute

[tex]K_f[/tex] = freezing point constant  

m = molality  of solution

We have :

[tex]K_f[/tex] of water = 1.86°C/m ,

Molality of solution = m = ?

[tex]KNO_3(aq)\rightarrow K^+(aq)+NO_3^{-}(aq)[/tex]

i = 2

Freezing point of solution = [tex]T_f=-14.5^oC[/tex]

Freezing point of pure water = T = 0°C

[tex]\Delta T_f=T-T_f[/tex]

[tex]\Delta T_f=0^oC-(-14.5 ^oC)=14.5^oC[/tex]

[tex]14.5^oC=2\times 1.86^oC\times m[/tex]

m = 3.898 molal

3.898 moles of potassium nitrate is dissolved in 1 kg of water or 1000 g of water.

Volume of water , V= 275 ml

Mass of water = m

Density of water= d = 1 g/mL

[tex]m=d\times V=1 g/ml\times 275 mL = 275 g[/tex]

Here, 3.898 moles of potassium nitrate is dissolved in 1 kg of water or 1000 g of water. Then moles of potassium nitarte present in 275 grams of water is :

[tex]\frac{3.989 mol}{1000}\times 275 =1.072 mol[/tex]

Mass of 1.072 moles of potassium nitrate :

1.072 mol × 101 g/mol = 108.27 g

108.27 grams of potassium nitarte must be added to 275 mL of water to produce a solution that freezes at -14.5°C.

2) Formula used an Elevation in boiling point;

[tex]\Delta T_b=T_b-T[/tex]

[tex]\Delta T_b=i\times K_b\times m[/tex]

where,

[tex]T_b[/tex] =boiling point of solution

T = boiling point of water

[tex]\Delta T_b[/tex] =Elevation in boiling point =  

i = van't Hoff factor of solute

[tex]K_b[/tex] = Boiling point constant  

m = molality  of solution

of the solution  

We have :

[tex]K_b[/tex] of water = 0.512°C/m ,

Molality of solution = m = 3.90 m

i =?

The boiling point of pure water = T = 100.00°C

The boiling point of solution = [tex]T_b[/tex]= 103.45°C

[tex]\Delta T_b=103.45^oC-100.00^oC=3.45^oC[/tex]

[tex]\Delta T_b=i\times K_b\times m[/tex]

[tex]3.45^oC=i\times 0.512 ^oC/m\times 3.90 m[/tex]

[tex]i=\frac{3.45^oC}{0.512 ^oC/m\times 3.90 m}=1.73[/tex]

1.73 is the actual value of the van't Hoff factor, i.

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