1. (5) From today\'s experiment, Sarah measured and recorded the following data:

Question

1. (5) From today's experiment, Sarah measured and recorded the following data: Barometric Pressure = 0.995 atmosphere; Boiling point of H2O = 99.8°C mass of flask+ septum: 98.3244 g mass of flask + septum+ unknown: 99.0834 g mass of flask + septum + water: 234.11 g density of water = 1.00 g/ml; From this data, calculate the molecular weight of the unknown. What must we assume about th behavior of the vapor? Explain why one must not use the analytical balance to weigh the flask septum filled with water. R = 0.0821 liter-atm/mole K: 1 liter 1000 ml

Explanation / Answer

1. Determining the molecular weight of the unknown

mass of water = 234.11 - 98.3244 = 135.7856 g

volume of flask = 135.7856 g/1 g/ml = 135.7856 ml = 0.136 L

Pressure = 0.995 atm

R = gas constant = 0.0821 L.atm/K.mol

T = temperature = 99.8 oC + 273 = 372.8 K

Using,

n = PV/RT

with n = moles of unknown

n = 0.995 x 0.135/0.0821 x 372.8 = 0.0044 mol

moles = grams/molecular weight

mass of unknown = 99.0834 - 98.3244 = 0.759 g

So,

molecular weight of the unknown = 0.759 g/0.0044 mol = 172.50 g/mol

The assumption made for the vapor is all the liquid is in the vapor state and is non-reactive to the septum and the flask. Analytical balances would not give the accurate mass of the flask with water and sepum when weighed and thus is not the preferred choice for this experiment.

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