A student following the procedure of this experiment obtained the following data

Question

A student following the procedure of this experiment obtained the following data for an unknown volatile liquid: mass of flak, boiling stone, foil cap, and unknown after cooling, g 83.350 mass of flak, boiling stone, and foil cap, g 82.657 water bath temperature, degree C 95.0 barometric pressure, in Hg 30.09 volume of the flask, mL 270 accepted molar mass of unknown, g mol^-1 86.2 Calculate the mass of the unknown. Express the water bath temperature in K. Express the barometric pressure in atmospheres.

Explanation / Answer

(a)

Given, Mass of flask + boiling stone + foil cap + unknown after cooling = 83.350 g = m1

Mass of flask + boiling stone + foil cap = 82.657 g = m2

Mass of Unknown = m1 - m2

=> 83.350 - 82.657 = 0.693 g

Mass of unknown = 0.693 g

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(b)

Given, the temperature of water bath in degree C = 95 degree C

Temperature in Kelvin = Temperature in degree C + 273.15

Temperature in K = 95 + 273.15

Temperature in Kelvin = 368.15 K

Therefore the water bath temperature in K = 368.15 K

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(c)

Given, barometric pressure = 30.09 Hg

We know that,

1 atm = 760 mm Hg

So, to convert 30.09 mm Hg into atm = 30.09 / 760 = 0.0396 atm

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