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

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.