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Determination of molecular weight of an unknown compound using an open end manometer.

Uploaded: 5 years ago
Contributor: Shriya Saha
Category: Physical Chemistry
Type: Solutions
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Filename:   22.docx (33.83 kB)
Page Count: 2
Credit Cost: 1
Views: 235
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Transcript
Q.22: An open tube manometer is connected to a flask containing 6.944 g of volatile compound with a total volume of 2.306 L of vapor at 22.06oC. The manometer tube is filled with water (d = 0.99777 g/mL); the level of water column in the open arm is 25.7 cm higher than in the arm connected to the flask. At the time of measurement, ambient pressure is 0.988 atm. Calculate the molar mass of the gas. (?Hg = 13.69 g/mL) Solution: Concept: The open tube manometer helps us in determining the pressure of an unknown gas which is being introduced from one end of the tube. If the pressure of the unknown gas is less than that of atmospheric pressure, then the height of the liquid in the column is more on arm connected to the unknown gas; and, if the pressure of the gas is higher than that of atmospheric pressure, then the height of the liquid in the column is more on the open end side. In this question, the pressure of the volatile compound is more than the atmospheric pressure as the height of liquid in the manometer is higher in the open arm of the tube. Once the pressure of gas is determined, we can determine the moles of the volatile compound present by using ideal gas law and then the molar mass. Since, the manometer is now filled with water instead of mercury, the height difference due to the pressure of the volatile compound and the atmospheric pressure can be determined in terms of mercury height difference by the densities of water and mercury. Calculations: 242887578422500The manometer can be shown below. Here, one end is attached to the flask containing the volatile compound and one end is open. Since the height of the water in the manometer is higher in the open end side, the pressure of the volatile compound is more than the atmospheric pressure. The atmospheric pressure is given in atm, which we need to convert to mm Hg. 0.988 atm × 760 mmHg1 atm=750.88 mmHg Height difference in terms of mercury hmercury=hwater× ?water?mercury=25.7 cm × 0.99777gmL13.69gmL=1.873 cm Hg =18.73 mm Hg Pressure of gas is calculated as Pgas= Patm+h in mm ?Pgas=750.88 mmHg+18.73 mm=769.61 mmHg This is the pressure of the volatile compound in the flask. We have been given the volume of vapor and the temperature at which forms vapor, we can calculate the moles of the compound present in the flask by ideal gas law PV=nRT. Pressure of vapor P=769.61 mmHg=769.61 mmHg×1 atm760 mmHg=1.0126 atm Volume of Vapor V=2.306 L Temperature T=22.06oC=22.06+273.15 K=295.21 K Universal Gas Constant R=0.08206 atm?Lmol?K Therefore, mole of the volatile compound present is n= PVRT= 1.0126 atm ×2.306 L0.08206 atm?Lmol?K ×295.21 K=0.09639 mol We know that moles=weightmolar mass ?molar mass=weightmoles Mass of the volatile compound present =6.944 g Hence, molar mass of the compound will be molar mass=weightmoles=6.944 g 0.09639 mol= 72.04067g/mol Therefore, molar mass of the volatile compound is 72.04 g/mol.

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