A 4.55 g sample of water is introduced into a 5.27 L flask containing some C_4 H

Question

A 4.55 g sample of water is introduced into a 5.27 L flask containing some C_4 H_10 gas. The flask is heated to 356.49 degree C at which temperature all of the water is converted to the gaseous phase, giving a total pressure in the flask of 11.134 atm. Calculate P_H_2 O (in atm) in the flask at 356.49 degree C. Report your answer to three decimal places in standard notation (i.e. 1.234 atm). Calculate P_C_4 H_10 (in atm) in the f,ask at 356.49 degree C. Report your answer to three decimal places in standard notation (i.e. 1.234 atm). i i ( SubrtlAnwwT) Tries 0/3 3. Calculate the mass of C_4 H_10 (in grams) in the flask at 356.49 degree C. Report your answer to three significant figures. Calculate X_C_4 H_10 in the flask at 356.49 degree C. Report your answer to three significant figures.

Explanation / Answer

1) moles of vapours formed = moles of water present

Mass of water = 4.55 grams

Moles = 4.55 / 18 = 0.252 moles

Total pressure of flask = 11.134 atm

Temperature = T = 356.49 +273 = 629.49 K

Pressure exerted by vapours = moles of vapour X R X T / volume = 0.252 X 0.0821 X 629.49 / 5.27 = 2.471 atm

So partial pressure = Pressure of vapours / total pressure = 0.222

2) Partial pressure of C4H10 = Pressure / total pressure

Pressure of C4H10 = 11.134 - 2.471 = 8.663 atm

Partial pressure = 8.663 / 11.134 = 0.778

3) Moles of C4H10 = mass / mol wt of C4H10

we know that

Moles = Pressure X volume / R X Temperature = 8.663 X 5.27 / 0.0821 X 629.49 = 0.883

So mass of C4H10 = 0.883 X molecular weight = 0.883 X 58 = 51.124 grams

4) Mole fraction of C4H10 = X C4H10 = moles of C4H10 / total moles

Total moles = 0.883 + 0.252 = 1.135

Mole fraction = 0.883 / 1.135 = 0.778

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