A gas of unknown molecular mass was allowed to effuse through a small opening under constant pressure conditions. It required 57 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 32 s for 1.0 L of O2 gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; that is, rate and time are inversely proportional.)

101.59 g/mol

Explanation:

Unknown gas = X

Using grahms's law of effusion;

RateX / RateO2  = √ (Molar massO2 / Molar massX)

RateX= Volume / time = 1 / 57 = 0.01754

RateO2 = Volume / Time = 1 / 32 = 0.03125

Molar massO2 = 32

Inserting the values into the equation;

0.01754 / 0.03125  =  √ (32 / Molar massX)

0.56128² = 32 / Molar massX

0.3150 * Molar massX = 32

Molar massX = 32 / 0.3150

Molar massX = 101.59 g/mol

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.

what is the question here. i think you messed up.