Origional Question: A mizture of helium and neon gases iscollected over water at

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Origional Question: A mizture of helium and neon gases iscollected over water at 28 degrees C and 745mmHg. If the partialpressure of helium is 368 mmHg, what is the partial pressure ofneon? (Vapor pressure of water at 28 degress C = 28.3 mmHg). Cramster says to add 745mmHg + 28.3 mmHg = 773.3 mm HG which is correct to get total pressure. Then it says to subract total from the neon which is likethis. 773.3mmHg - 368mmHg It says that the answer is 349mmHg but I get 405 mmHg I dunno what to do here Origional Question: A mizture of helium and neon gases iscollected over water at 28 degrees C and 745mmHg. If the partialpressure of helium is 368 mmHg, what is the partial pressure ofneon? (Vapor pressure of water at 28 degress C = 28.3 mmHg). Cramster says to add 745mmHg + 28.3 mmHg = 773.3 mm HG which is correct to get total pressure. Then it says to subract total from the neon which is likethis. 773.3mmHg - 368mmHg It says that the answer is 349mmHg but I get 405 mmHg I dunno what to do here

Explanation / Answer

The given data is

           The pressure of mixture (He+Ne+ ) is 745 mmHg at

           The partial pressure of the Helium is 368 mmHg

           And Thevapour pressure of water is 28.3 mmHg

           The partial pressure of the Neon =?

If we substract the sum of the partial pressures of heliumand water from the Total   pressure we get partialpressure of the Neon .

The partial pressure of the Neon = 745 mmHg-(368 mmHg+28.3mmHg)

                                                      = 745 mmHg- 396.3 mmHg

                                                      = 348.7 mmHg

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