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In the 19th century, J. B. A. Dumas devised a method for finding the molar mass of a volatile liquid from the volume, temperature, pressure, and mass of its vapor (next column). He placed a sample of such a liquid in a flask that was closed with a stopper fitted with a narrow tube, immersed the flask in a hot water bath to vaporize the liquid, and then cooled the flask. Find the molar mass of a volatile liquid from the following:
Mass of empty flask = 65.347 g
Mass of flask filled with water at 25° C = 327.4 g
Density of water at 25°C = 0.997 g/mL
Mass of flask plus condensed unknown liquid = 65.739 g
Barometric pressure = 101.2 kPa
Temperature of water bath 5 99.8°C

Respuesta :

The molar mass of a volatile liquid is 41.94 g/ mol.

Multiply the molar mass (from the periodic desk) of each detail by means of the variety of atoms of that element gift inside the compound. 3. upload all of it collectively and placed devices of grams/mole after the quantity.

In chemistry, the molar mass of a chemical compound is defined as the mass of a pattern of that compound divided by the amount of substance that's the number of moles in that pattern, measured in moles. The molar mass is a bulk, no longer molecular, an asset of a substance.

The molar mass is the mass of all the atoms in a molecule in grams in step with mole. To calculate the molar mass of a molecule, we first obtain the atomic weights from the man or woman elements in a periodic desk. We then depend on the wide variety of atoms and multiply it by way of the person's atomic hundreds.

Mass of Empty Flask = 65.3479

Mass of Flask with water

327-49

density of water 0-997 g/mal

mass of Flask + condensing unknown liquid = 65-7599

Mass of condensing unknown liquid = (65-789-69-347)g

=0.3929

Pressure 109.2 KPO

Temperature

= 95+1°C

Mass of water =(327-4-65-347)g

262-0539

d = m/v

0-9972-262-0539

V

V=262.84ml

V = 2621

Rmezlice

109.2 KPa

=109-2 x 1000 Pa

=109200 Po

Pressure in atm

Pressure- 109200 atm

101325

=1.078 atm

Temperature 95-1°C

= 95+1 +273 K =368° 1 K

using PV = nRT

16078am X 0.262 -0-392 9 x 0-08-21 tan mal x 368-1 K

M

M =0.3929 x0-0821 bad" x 568-1

1.078 x 262

M- 41.94 g/mol

Molex weight of unknown liquid =41.949/mol

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