n this experiment, what measurement will limit the number of significant figures

Question

n this experiment, what measurement will limit the number of significant figures? 2. A gaseous hydrocarbon collected over water at a temperature of 21 °C and barometrie pressure of 753 torr occupied a volume of 48.1 mL. The hydrocarbon in this volume weighs 0.1133 g. Calculate the molar mass of the hydrocarbon. The vapor pressure of water at 21 °C is 18.7 torr. 3. Using the experimental data from question #2, how would your result be affected ifyou do not account for the vapor pressure of water (assume vapor pressure of water is negligible)? Using the experimental data from question #2, calculate the volume of the hydrocarbon at standard temperature and pressure (STP). (Note: the vapor pressure of water at STPis 4.6 torr). 4.

Explanation / Answer

1. Volume should limit the significant figures,as it has only 3 significant figures,while others have either 3 or more than 3

T = 21 oC = 273.15 + 21 = 294.15 K

1 atm = 760 torr

753 Torr = 753 / 760 atm = 0.991 atm

V = 48.1 mL = 0.0481 L

partial pressure of H2 = PT - PH2O = 753 torr - 18.7 torr = 734.3 torr

= 734.3 torr / 760 torr = 0.966 atm

n = PV/RT = 0.966 atm x 0.0481 L / (0.08206 atm.L/mol.K x 294.15 K) = 0.00192631 mol

mass / moles = molar mass

0.1133 g / 0.00192631 mol = 58.8 g/mol

3.

If water pressure is taken negligible

n = PV/RT = 0.991 atm x 0.0481 L / 0.08206 atm.L/mol.K x 294.15 K = 0.00197578 mol

molar mass = mass / moles = 0.1133 g/ 0.00197578 mol = 57.3 g/mol

4.

at STP

pressure = 760 torr

water vapour pressure = 4.6 torr

Hydrogen partial pressure = 760 torr - 4.6 torr = 755.4 torr

755.4 torr / 760 torr = 0.993947 atm

T = 273 K

V = nRT/P = 0.00192631 mol x 0.082060 atm.L/mol.K x 273K / 0.993947 atm = 0.0434 L = 43.4 mL

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