Question

Suppose that 0.00150 moles of CO2 (44.0 g/mol) effuse out of a pinhole in 1.00 hour. How many moles of N2 (28.0 g/mol) would effuse out of the same pinhole in 1.00 hour? Assume both gases are at the same temperature and pressure.

Answer 1

Answer:

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Explanation:

Answer 2

Answer:

0.0019mol

Explanation:

To solve this problem, first we calculate for the rates of the effusion:

For CO2,

Number of mole = 0.00150 mol

Time = 1h

R1 = mol / time

R1 = 0.0015/1

R1 = 0.0015mol/h

Molar Mass of CO2 (M1) = 44.0 g/mol

For N2,

R2 =?

Molar Mass of N2 (M2) = 28.0 g/mol

Using Graham's law, we can obtain Rate (R2) of N2 as follows:

R1/R2 = √(M2/M1)

0.0015/R2 = √(28/44)

0.0015/R2 = 0.7977

R2 = 0.0015 /0.7977

R2 = 0.0019mol/h

But rate = mole /time

Time = 1h

Rate od N2 = 0.0019mol/h

0.0019 = mole /1

Mole = 0.0019mol

Therefore, the number of mole of N2 is 0.0019mol