1) What is the equilibrium partial pressure of water vapor above a mixture of 62
ID: 948800 • Letter: 1
Question
1) What is the equilibrium partial pressure of water vapor above a mixture of 62.2 g H2O and 35.2 g HOCH2CH2OH at 55 °C. The partial pressure of pure water at 55.0 °C is 118.0 mm Hg. Assume ideal behavior for the solution.
2
Which of the following statements is INCORRECT?
a)The solubility of a gas in water decreases with increasing temperature.
b)The dissolution of a gas in water is usually an exothermic process.
c)The solubility of a gas in water is proportional to the partial pressure of the gas above the water.
d)The solubility of a gas in water is inversely proportional to the molar mass of the gas.
e)The relationship between the solubility of a gas and its partial pressure is known as Henry's law.
a)The solubility of a gas in water decreases with increasing temperature.
b)The dissolution of a gas in water is usually an exothermic process.
c)The solubility of a gas in water is proportional to the partial pressure of the gas above the water.
d)The solubility of a gas in water is inversely proportional to the molar mass of the gas.
e)The relationship between the solubility of a gas and its partial pressure is known as Henry's law.
Explanation / Answer
Moles of water = mass/molar mass of H2O = 62.2 /18.015 = 3.4527
Moles of ethlene glycol = 35.2 /62.07 = 0.567
mol fraction of water = moles of water / ( total moles of mixture) = ( 3.4527 /3.4527+0.567) = 0.859
Partial pressure of H2O = mol fraction of H2O x vapor pressure of Pure H2O = 0.859 x 118 = 101.36 mm Hg
2) Solubility of gas increases as Molarmass of gas increases since bigger gas molecules will have more interactions and hence molre solubility
Hence statement D is wrong
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