- ) An open-ended manometer is connected to a sample of pure argon gas in a 250.0 mL flask. Atmospheric pressure is 758.2 mmHg at 24 °C. The manometer’s mercury in the arm attached to the gas is 15.4 mmHg lower than in the arm open to the atmosphere.

- Draw a picture of the manometer/flask set up in question #3 and explain how to determine the pressure of the argon gas in the flask.
- How many grams of argon are in the flask?

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Following is the -complete Answer -&- Explanation: for the given Question, in....typed and image format.... =>Answer: - Part:(a): Pressure of Argon gas (Ar) in the flask: Pgas = 773.6 torr  - Part:(b): Mass of Argon in the flask: = 0.399 g (grams) of Argon.. =>Explanation: Following is the complete Explanation, for the above Answer.... tGiven: - Volume of the flask containing argon gas: V = 250.0 mL x (1.0 L / 1000 mL) = 0.25 L (Liters) - Atmospheric pressure: Patm = 758.2 mmHg (milli-meters of mercury, Hg) - Tempearture of the flask containing Argon gas: T = 24oC = (273.15 + 24 )K = 297.15 K (Kelvin) - Manometer's mercury in the arm attached to the gas is 15.4 mmHg ,lower than in the arm open to the atmosphere. =>Asked:  (a) Draw a picture of the manometer/ flask, set up in the question and explain how one can determine the pressure of the argon gas in the flask. (b) Determine how many grams of Argon (Ar) gas are in the flask. tStep - 1: u200bu200bu200bu200bu200bu200bu200bFollowing is the drawn picture of a manometer/flask set up, as desired in the question, in....image format.... Pars Patn (afmospheric prassure).\int v^{3}-Dren - End Manometer tStep -2: u200bu200bu200bu200bu200bu200bu200bAs we can see in the above picture, since the mercury (Hg) in the arm attached to the flask containing Argon( Ar ), is lower than the arm open to the atmosphere, we can say that the pressure of the gas Argon (Ar) must be more than Amospheric pressure, and gas pressure in this open end manometer, overcomes the atmospheric pressure and further raises the mercury through 'h' mmHg. Thus, the pressure of the gas (Pgas) will be the following: => Pgas = Patm + h mmHg  ---------------------------------Eq. (1) tStep - 3: u200bu200bu200bu200bu200bu200bu200bPlugging in values in the above Eq.(1), we will get the following: => Pgas = Patm + h mmHg  ---------------------------------Eq. (1) => Pgas = Patm + h mmHg  Plugging in values.... => Pgas = (758.2 + 15.4 ) mmHg = 773.6 mmHg = 773.6 torr  Pressure of Argon (Ar ) in the flask: Pgas = 773.6 torr  tStep - 4: u200bu200bu200bu200bu200bu200bu200bWe know Argon (Ar) can be regarded as an Ideal Gas, and thus the following Ideal Gas law would be applicable to Argon as well.... => Ideal Gas Law: PV = nRT --------------------------------------Eq.(2) Where: - P = pressure of the gas = 773.6 torr - V = Volume of the gas = 0.250 L (liters) - n = Number of moles of the gas - R = Ideal Gas constant = 62.36 L. torr. K-1. mol-1    - T = Temperature of the gas = 297.15 K (Kelvin) tStep - 5: From the above Eq.(2), we can obtain the following equation to determine the number of moles, 'n'  => n = PV/ RT -----------------------------------Eq. (3) Plugging in values... => n = (773.6 torr x 0.25 L)/ (62.36 L. torr. K-1. mol-1 x 297.15 K) = 0.01 mol (moles) of Argon (Ar) gas tStep - 6: u200bu200bu200bu200bu200bu200bu200bWe know,  =>Molar mass of Argon (Ar) = 39.948 g/mol (i.e. grams per mole) Therefore, => Mass of Argon in the flask: = (0.01 mol Ar) x (39.948 g/ mol Ar) = 0.399 g (grams) of Argon..   --------------------------------------------------------------------------------------------------------         ...