6/1712017 11:55 PM 6/11/2017 07:55 PM Gradebook Periodic Table estion 16 of 31 M

Question

6/1712017 11:55 PM 6/11/2017 07:55 PM Gradebook Periodic Table estion 16 of 31 Mapoteo A Sapling Learning Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 5.95 mL of O2 had passed through the membrane, but only 3.00 mL of of the unknown gas had passed through. What is the molar mass of unknown Number g/ mol o Assig Availab Due Da Points Grade Descrip Policies You car solution You can O eText O Help O Web O Techn

Explanation / Answer

For solving this, We make use of Graham's Law:

r1 / r2 = SQRT(MM2 / MM1)

Where MM1= molar mass of unknown gas

MM2 = molar mass of O2

Let certain amount of time = 1minute.

Therefore:

r1 = 5.95 mL/min ( O2)
r2 = 3.00 mL/min (unknown gas)

5.95 / 3.00 = SQRT (x / 32.0)

1.983 = SQRT (x /32.0)

Now squaring both sides:

3.93 = x / 32.0

x = 125.76 g/mol

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