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swat32
3 years ago
5

A supersaturated solution can be prepared by dissolving solute in solvent while adding ______. A supersaturated solution contain

s more ______ than can ordinarily be dissolved in the solvent at room temperature. A solution may remain supersaturated until _______ is initiated, often by adding solid to the solution or by allowing solvent to evaporate.
Chemistry
1 answer:
Delicious77 [7]3 years ago
4 0

Answer:

Heat, solutes and high temperature.

Explanation:

A supersaturated solution can be formed by dissolving solute more solute in solvent by increasing temperature of the solution. A supersaturated solution contains more quantity of solutes than can be dissolved in the solvent at room temperature. A solution may remain supersaturated until the solution has high temperature and when the temperature started lower, the extra dissolve solutes begin undissolved and remain suspended in the solution.

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Answer:

Mole fraction of alcohols in liquid phase x_1=0.2727\& x_2=0.7273.

Mole fraction of alcohols in vapor phase y_1=0.1468\& y_2=0.8516.

Explanation:

The total vapor pressure of the solution = p =38.6 Torr

Partial vapor pressure of the  n-propyl alcohol =p^{o}_1=21.0 Torr

Partial vapor pressure of the isopropyl alcohol =p^{o}_2=45.2 Torr

p=x_1\times p^{o}_1+x_2\times p^{o}_2  (Raoult's Law)

p=x_1\times p^{o}_1+(1-x_1)\times p^{o}_2

38.6 Torr=x_1\times 21.0 Torr+(1-x_1)\times 45.2 Torr

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x_2=1-0.2727=0.7273

x_1\& x_2 is mole fraction in liquid phase.

Mole fraction of components in vapor phase y_1\& y_2

p_1=y_1\times p (Dalton's law of partial pressure)

y_1=\frac{p_1}{38.6 Torr}=\frac{p^{o}_1\times x_1}{38.6 Torr}

y_1=\frac{21.0 Torr\times 0.2727}{38.6 Torr}=0.1468

y_1=\frac{p_2}{38.6 Torr}=\frac{p^{o}_2\times x_2}{38.6 Torr}

y_2=\frac{45.2 Torr\times 0.7273}{38.6 Torr}=0.8516

Mole fraction of alcohols in vapor phase y_1=0.1468\& y_2=0.8516

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