A student weighed an empty 250mL Erlenmeyer flask and rubber stopper; their comb

Question

A student weighed an empty 250mL Erlenmeyer flask and rubber stopper; their combined weight was 87.35g. Precisely 3.00g of an unknown liquid was added to the flask. As in this experiment, the flask was heated for several minutes in a hot water bath until the temperature in the flask became constant at 77.6 degrees C (the temp of the surrounding water bath was 82.3C). The atmospheric pressure in the lab was measured at 755.8 mm Hg. After heating, the flask was quickly placed in an ice/water bath, re-stoppered, carefully dried off, and reweighed; the weight of the flask, stopper, and contents was now 88.52g. The flask was rinsed with water and completely filled with distilled water. The weight of the flask, stopper, and water is now 364.23g.

What is the MM of this unknown liquid? The unknown liquid was actually chloroform, CHCl3. What is the % error for the MM obtained in this experiment?

Explanation / Answer

Solution:- Given data is...

mass of empty flask and rubber stopper = 87.35g

mass of unknown liquid taken = 3.00g

Temperature of flask on heating = 77.6 dgree C = 77.6 + 273 = 350.6K

atmospheric pressure in lab = 755.8 mm Hg = 755.8 mm Hg x 1 atm/760 mm Hg = 0.9945 atm

new weight of flask, stopper and content = 88.52g

weight of flask, stopper and distilled water = 364.23g

Now we could calculate the mass of the vapor which fills the flask and it is the difference of 88.52g and 87.35g.

mass of vapor = 88.52 g - 87.35g = 1.17g

mass of distilled water = 364.23g - 87.35g = 276.88g

density of water is 1g/ml.

so, the volume of the flask = 276.88g x 1ml/1g = 276.88ml = 0.27688 L

we know that, PV = nRT

let's plug in the values in it...

0.9945 x 0.27688 = n x 0.0821 x 350.6

n = 0.9945 x 0.27688 / 0.0821 x 350.6

n = 0.009566 mole

now we have mass and molar mass. So, molar mass could easily be calculated.

molar mass = mass/moles = 1.17g/ 0.009566 mole = 122.31 g/mole

So, the molar mass of the unknown is 122.31 g/mole.

percent error in molar mass(MM) = [(experimental value - accepted value)/accepted value] x 100

accepted molar mass of CHCl3 is 119.38 g/mole. So...

= [(122.31 - 119.38)/119.38] x 100

= 2.45%

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