96.0 g. of a gas occupies 48.0 L at 700.0 mm Hg and 20.0 °C. What is its molecular weight?

Molecular weight of the gas is 52.2 g/mol.

Ideal gas equation is a hypothetical equation which tells about the behavior of the gas under standard condition and it ca be described as:

PV = nRT, where

P = pressure = 700.0 mm Hg (given)

V = volume = 48.0 L (given)

R = universal gas constant = 0.08206 L atm mol¯¹ K¯¹

T = temperature = 20.0 °C = 293 K

On putting theses values in the above equation we get for moles is as:

n = (700 × 48) / (0.08206 × 293) = 1.8 moles

We know that moles can be represented as:

n = W / M, where

W = given weight = 96g

M = molecular weight = to find?

Putting values in the mole equation, we get

M = 96 / 1.8 = 52.2 g/mol

Hence, molecular weight of gas is 52.2 g/mol.

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I hope this helps 52.2 g/mol

Explanation:

1) Solve for the moles using PV = nRT:

n = PV / RT

n = [(700.0 mmHg / 760.0 mmHg atm¯1) (48.0 L)] / [(0.08206 L atm mol¯1 K¯1) (293.0 K)]

n = 1.8388 mol

2) Divide the grams given (96.0) by the moles just calculated above:

96.0 g / 1.8388 mol = 52.2 g/mol