A cigarette lighter weighs 16.3482 g initially. This lighter is opened under wat

Question

A cigarette lighter weighs 16.3482 g initially. This lighter is opened under water and 12.34 ml of gas bubbling from the lighter is collected by water displacement in an inverted burette. The lighter is dried and re-weighed to give a final mass reading of 16.3195 g. The barometric pressure when the gas is collected is 754.6 mm Hg (or torr). The final equilibrated temperatur if measured to be 21.2 degree Celsius. What is the molar mass of the gas that was inside the lighter? Show your work. (Hint: Might need to look up the vapor pressure of water since it is present in the burette along with the lighter gas.)

Explanation / Answer

Mass of gas is 16.3482-16.3195 == 0.0287 g

Volume of gas = 12.34 ml = 0.01234 L

Pressure of gas = total pressure - vapor pressure of water at 21.2 oC

= 754.6 mm Hg -18.892 mm Hg

= 735.7 mm Hg

PV = n RT

n =PV/RT

n =(735.7 mm Hg /760 mm Hg ) x 0.01234 L / ( 0.082 x 294.35 K )

= 0.000494 moles

moles = mass / molar mass

molar mass = mass / moles

= 0.0287 g / 0.000494 moles

= 57.98 g/mol

= 58 g/mol

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.