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.

Answers

Answer 1

Answer:

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 ;

$\Delta T_f=T-T_f$

$\Delta T_f=i* K_f* m$

where,

$T_f$ =Freezing point of solution

T = Freezing point of water

$\Delta T_f$ =depression in freezing point =

i = van't Hoff factor of solute

$K_f$ = freezing point constant

m = molality of solution

We have :

$K_f$ of water = 1.86°C/m ,

Molality of solution = m = ?

$KNO_3(aq)\rightarrow K^+(aq)+NO_3^(-)(aq)$

i = 2

Freezing point of solution = $T_f=-14.5^oC$

Freezing point of pure water = T = 0°C

$\Delta T_f=T-T_f$

$\Delta T_f=0^oC-(-14.5 ^oC)=14.5^oC$

$14.5^oC=2* 1.86^oC* m$

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

$m=d* V=1 g/ml* 275 mL = 275 g$

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 :

$(3.989 mol)/(1000)* 275 =1.072 mol$

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;

$\Delta T_b=T_b-T$

$\Delta T_b=i* K_b* m$

where,

$T_b$ =boiling point of solution

T = boiling point of water

$\Delta T_b$ =Elevation in boiling point =

i = van't Hoff factor of solute

$K_b$ = Boiling point constant

m = molality of solution

of the solution

We have :

$K_b$ 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 = $T_b$ = 103.45°C

$\Delta T_b=103.45^oC-100.00^oC=3.45^oC$

$\Delta T_b=i* K_b* m$

$3.45^oC=i* 0.512 ^oC/m* 3.90 m$

$i=(3.45^oC)/(0.512 ^oC/m* 3.90 m)=1.73$

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

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Explanation:

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Answer:

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Answers

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Explanation:

Answer:

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Explanation:

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Answers

Answer:

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Explanation:

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Answers

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Answers

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Answers

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depends how many sig figs you are rounding to so I won't round to tenth or hundredth but the answer is 2.037 * 10^7

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