A gas of unknown molar mass was allowed to effuse through a small opening under constant pressure conditions. It required 1130 s for 1.00 L of this gas to effuse. Under identical experimental conditions it required 600. s for 1.00 L of ozone (03) 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; that is, rate and time are inversely proportional.) O a) The molar mass is 13.5 g/mol. O b) The molar mass is 65.9 g/mol. O c) The molar mass is 170.3 g/mol. O d) The molar mass is 90.4 g/mol. O e) The molar mass is 35.0 g/mol.

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Chapter1: Chemical Foundations
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A gas of unknown molar mass was allowed to effuse through a small opening under constant pressure conditions. It required 1130 s for 1.00 L of this gas to effuse. Under identical
experimental conditions it required 600. s for 1.00 L of ozone (03) 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; that is, rate and time are inversely proportional.)
O a) The molar mass is 13.5 g/mol.
O b) The molar mass is 65.9 g/mol.
O c) The molar mass is 170.3 g/mol.
O d) The molar mass is 90.4 g/mol.
O e) The molar mass is 35.0 g/mol.
Transcribed Image Text:A gas of unknown molar mass was allowed to effuse through a small opening under constant pressure conditions. It required 1130 s for 1.00 L of this gas to effuse. Under identical experimental conditions it required 600. s for 1.00 L of ozone (03) 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; that is, rate and time are inversely proportional.) O a) The molar mass is 13.5 g/mol. O b) The molar mass is 65.9 g/mol. O c) The molar mass is 170.3 g/mol. O d) The molar mass is 90.4 g/mol. O e) The molar mass is 35.0 g/mol.
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