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dem82 [27]
3 years ago
9

Which solution has the highest vapor pressure?

Chemistry
1 answer:
nata0808 [166]3 years ago
7 0

Answer:

the highest vapor pressure is the solution (b) P = 23.552 torr

∴ sln (a): P = 23.33 torr

∴ sln (c): P = 23.370 torr

Explanation:

vapor pressure:

  • ΔP = - (P*a)(Xb,l) = P - P*a

∴ a: solvent (water)

∴ b: solute

assuming T=25°C:

∴ P*a (25°C) = 23.8 torr

molecular weight:

∴ Mw H2O = 18 g/mol

∴ Mw glucose: 180.156 g/mol

∴ Mw sucrose = 342.3 g/mol

∴ Mw potassium acetate = 98.15 g/mol

∴ mol H2O = (100 mL)(1 g/mL)(mol/18 g) = 5.55 mol H2O

a) moles glucose = (20.0 g)(mol/180.156 g) = 0.111 mol

⇒ Xb,l = (0.111 mol C6H12O6)/(5.55 + 0.111) = 0.0196

⇒ ΔP = - (23.8 torr)(0.0196) = - 0.466 torr = P - P*a

⇒ P = 23.8 torr - 0.466 torr = 23.33 torr

b) moles sucrose = (20 g)(mol/342.3 g) = 0.0584 mol

⇒Xb,l = (0.0584)/(5.55 + 0.0584) = 0.0104

⇒ ΔP = - (23.8 torr)(0.0104) = - 0.248 torr

⇒ P = 23.8 torr - 0.248 torr = 23.552 torr

c) moles KC2H3O2 = (10 g)(mol/98.15 g) = 0.102 mol

⇒ Xb,l = (0.102)/(5.55 + 0.102) = 0.0180

⇒ ΔP = - (23.8 torr)(0.0180) = - 0.429 torr

⇒ P = 23.8 torr - 0.429 torr = 23.370 torr

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What is the density of 4 .8 grams of Oxygen gas at STP?
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Answer:

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Explanation:

The density of a gas can be obtained using the gas ideal equation and the molar mass of the gas.

This is the decution of the final formula:

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Now, you just need to substitute values:

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As you see, I have not used the 4.8 grams datum. That is because the density of the gases may be calculated from the temperature, pressure and molar mass of the gas, using the ideal gas equation.

Since, you have the mass of gas, you might use this other procedure:

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  • Density = mass / volume = 4.8 g / 3.36 liter = 1.4 g/liter (same result)

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