Question

In: Chemistry

A mixture of gases collected over water at 14oC has a total pressure of 1.198 atm...

A mixture of gases collected over water at 14oC has a total pressure of 1.198 atm and occupies 72 mL. How many grams of water escaped into the vapor phase?

Solutions

Expert Solution

The total pressure of gases over water at140 C, P = 1.198 atm

at 140 C the vapour pressure ofwater = 0.016atm

                 Total volume = 72.mL

                                     = 0.072 L

             Temperature   = (14 +273 ) K

                                      = 287 K

              Ideal gas equation PV = nRT

               so number of water moles , n = PV / RT

                                                              = ( 0.016 atm * 0.072L) / (0.08206 L.atm /mol.K*287K)

                                                              = 4.89 x 10-5 mol

                molar mass of water = 18 g / mol

          Therefore mass of water present in vapor state ,m = 18g /mol x 4.89 x 10-5 mol

                                                                                      = 8.80 x 10-4 g

                          8.80 x 10-4 g of water is escaps into vapor phase


Related Solutions

PART A A mixture of He, Ar, and Xe has a total pressure of 2.20 atm...
PART A A mixture of He, Ar, and Xe has a total pressure of 2.20 atm . The partial pressure of He is 0.400 atm , and the partial pressure of Ar is 0.450 atm . What is the partial pressure of Xe? Express your answer to three significant figures and include the appropriate units. PART B A volume of 18.0 L contains a mixture of 0.250 mole N2 , 0.250 mole O2 , and an unknown quantity of He....
Part A A mixture of He, Ar, and Xe has a total pressure of 2.70 atm...
Part A A mixture of He, Ar, and Xe has a total pressure of 2.70 atm . The partial pressure of He is 0.450 atm , and the partial pressure of Ar is 0.450 atm . What is the partial pressure of Xe? Express your answer to three significant figures and include the appropriate units. Part B A volume of 18.0 L contains a mixture of 0.250 mole N2 , 0.250 mole O2 , and an unknown quantity of He....
A mixture of CH4 (g) and C2H6 (g) has a total pressure of 0.53 atm. Just...
A mixture of CH4 (g) and C2H6 (g) has a total pressure of 0.53 atm. Just enough O2 was added to the mixture to bring about it's complete combustion to CO2 (g) and H2O (g). The total pressure of the two product gases is found to be 2.2 atm. Assuming constant volume and temperature, find the mole fraction of CH4 in the original mixture.
A mixture of CH4 (g) and C2H6 (g) has a total pressure of 0.53 atm. Just...
A mixture of CH4 (g) and C2H6 (g) has a total pressure of 0.53 atm. Just enough O2 was added to the mixture to bring about it's complete combustion to CO2 (g) and H2O (g). The total pressure of the two product gases is found to be 2.2 atm. Assuming constant volume and temperature, find the mole fraction of CH4 in the original mixture.
Dalton's law states that the total pressure, Ptotal, of a mixture of gases in a container...
Dalton's law states that the total pressure, Ptotal, of a mixture of gases in a container equals the sum of the pressures of each individual gas: Ptotal=P1+P2+P3+… The partial pressure of the first component, P1, is equal to the mole fraction of this component, X1, times the total pressure of the mixture: P1=X1×Ptotal The mole fraction, X, represents the concentration of the component in the gas mixture, so X1=moles of component 1total moles in mixture Part A Three gases (8.00...
Dalton's law states that the total pressure, Ptotal, of a mixture of gases in a container...
Dalton's law states that the total pressure, Ptotal, of a mixture of gases in a container equals the sum of the pressures of each individual gas: Ptotal=P1+P2+P3+… The partial pressure of the first component, P1, is equal to the mole fraction of this component, X1, times the total pressure of the mixture: P1=X1×Ptotal The mole fraction, X, represents the concentration of the component in the gas mixture, so X1=moles of component 1total moles in mixture Question 1 ---- Three gases...
1.The total pressure of a mixture of O2(g) and H2(g) is 1.95 atm. The mixture is...
1.The total pressure of a mixture of O2(g) and H2(g) is 1.95 atm. The mixture is ignited and the resulting water is removed. The remaining mixture is pure H2(g) and exerts a pressure of 0.210atm when measured at the same temperature and volume as the original mixture. What were the mole fractions of O2(g) and H2(g) in the original mixture?
1.) The total pressure (PT) os a sample of a gas collected over water at 13.0oC...
1.) The total pressure (PT) os a sample of a gas collected over water at 13.0oC is 0.588 atm. Determine the partial pressure (Pi) of the gas. 2.) Rearrange the ideal gas equation to solve for density (g/v). Hint: you will need to make a substitution for moles "n" 3.) Use your quation from (z) and the result from (l) to calculate the molecular weight of a gas if 303 ml of the gas, collected at 28oC, weighed 0.347g.
1a/ A mixture of helium and carbon dioxide gases, at a total pressure of 743 mm...
1a/ A mixture of helium and carbon dioxide gases, at a total pressure of 743 mm Hg, contains 0.397 grams of helium and 9.33 grams of carbon dioxide. What is the partial pressure of each gas in the mixture? PHe = (......) mm Hg PCO2 = (......) mm Hg 1b/ A mixture of methane and xenon gases contains methane at a partial pressure of 488 mm Hg and xenon at a partial pressure of 386 mm Hg. What is the...
1. A mixture of hydrogen and oxygen gases, at a total pressure of 874 mm Hg,...
1. A mixture of hydrogen and oxygen gases, at a total pressure of 874 mm Hg, contains 0.316 grams of hydrogen and 8.78 grams of oxygen. What is the partial pressure of each gas in the mixture? PH2 =  mm Hg PO2 =  mm Hg 2. A mixture of carbon dioxide and hydrogen gases contains carbon dioxide at a partial pressure of 279 mm Hg and hydrogen at a partial pressure of 477 mm Hg. What is the mole fraction of each...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT