20.0 g of an unknown gas is combined with helium in a 1500. mL container at 25 d

Question

20.0 g of an unknown gas is combined with helium in a 1500. mL container at 25 degree C giving a total pressure of 86.110 atm. Given that the partial pressure of helium is 81.513 atm, what is the molar mass of the unknown gas? A sample of H_2 gas that was collected over water occupies a volume of 47.1 mL at 23 degree C and 744 mm Hg. P_H_2 O = 21.1 mm Hg at 23 degree Celsius. What volume in mL would the H_2 gas occupy at STP conditions? According to the postulates of Kinetic Molecular Theory, the root-mean-square (rms) speed of the molecules of a given gas is proportional to the: A. Reciprocal of the absolute temperature B. Celsius temperature squared C. squareroot of the absolute temperature D. Absolute temperature E. Absolute temperature squared

Explanation / Answer

1.Total pressure(P) = partial pressure of unknown + partial pressure of He (pHe)

So partial pressure of unknown , p = P - pHe

= 86.110 - 81.513

= 4.597 atm

Calculation of number of moles of unknown :

We know that ideal gas equation is PV = nRT

Where

T = Temperature = 25oC = 25+273 = 298 K

P = pressure = 4.597 atm

n = No . of moles = ?

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = volume of the container = 1500 mL = 1.5 L

plug the values we get n = (PV) / (RT)

= 0.282 moles

We know number of moles , n = mass/molar mass

So Molar mass of unknown = mass/number of moles of unknown

= 20.0 g / 0.282 mol

= 71.0 g/mol

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