Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

The mass of the Florence flask + stopper + air used in the molecular weight dete

ID: 304121 • Letter: T

Question

The mass of the Florence flask + stopper + air used in the molecular weight determination experiment was 116.817 g. Si Florence flask. A one-hole stopper was placed on the flask water bath. As soon as all the unknown liquid was vapor x milliliters of an unknown volatile liquid was transferred into the and the flask was then placed in a hot 56 e one-hole stopper was taken out of the 94°C water bath and the one-hole e flask + stopper + unknown gas ized over a barometric pressure of 7 stopper was replaced with a stopper. The mass of the Florenc (after cooling) was found to be 117.293 g. At the end of the experiment, the Florence flask was filled with water and its volume was found to be 272. mL. a) What is the mass (in g) of the air (d = 1.18 g/L) in the Florence flask? (5 pts) b) What is the mass (in g) of the unknown gas? (5 pts) c) What is the molecular weight (in g/mol) of the unknown? (10 pts)

Explanation / Answer

a) In the last line volume occupied by water given = 272 mL which is equal to volume of flask and which will be equal to volume occupied by air in the starting of experiment.

Therefore volume of air = 272 mL and d = 1.18 g/L (given)

d = m/V

1.18 g/L = mass of air (g)/0.272 L

On solving this mass of air comes out to be = 0.32096 g

b) flask + stopper + air mass = 116.817 (given)

mass of air = 0.32096 (calculated)

So, Flask + stopper mass = 116.817 - 0.32096 = 116.496 g

Flask + stopper + unknown gas mass = 117.293 g

Therefore, mass of unknown gas = 117.293 - 116.496 = 0.797 g

c) use ideal equation PV = nRT to find no. of moles of unknown gas

P = 756 torr

As 760 torrr = 1 atm So, 756 torr = 756/760 atm

Volume of flask = 272 mL = 0.272 L

R = 0.0821 Latm/Kmol

T = 94 °C = 367 K

Therefore using PV = nRT

no of moles of gas (n) = 756 * 0.272/760 * 0.0821 * 367

no of moles comes out to be = 8.98 * 10-3 mole

mass of gas claculated = 0.797 g

no of mole = given mass / molecular mass

So, molecular weight of gas = mass of gas/moles of gas

Mol. wt. = 0.797/ 8.98 * 10-3

mol wt = 88.75 g/mol