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For a 80- g sample of fused copper catalyst, a volume of 7.6×103 mm3 of nitrogen (measured at standard temperature and pressure, 0 ∘c and 1 atm ) is required to form a monolayer upon condensation. calculate the surface area of the catalyst. (take the area covered by a nitrogen molecule as 0.162 nm2 and recall that, for an ideal gas, pv=nrt , where n is the number of moles of the gas.)

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Answer:

First, we need to calculate the number of moles of nitrogen gas required to form a monolayer:

n = (pv) / (rt)

where p is the pressure, v is the volume, r is the ideal gas constant, and t is the temperature in Kelvin.

At standard temperature and pressure, we have:

p = 1 atm

v = 7.6×10^3 mm^3 = 7.6×10^-6 m^3

t = 273 K

r = 8.31 J/(mol K)

So, n = (1 atm x 7.6×10^-6 m^3) / (8.31 J/(mol K) x 273 K) = 3.13×10^-7 mol

Next, we can calculate the number of nitrogen molecules in this amount of gas:

N = n x Na

where Na is Avogadro's number (6.02×10^23 molecules/mol).

N = 3.13×10^-7 mol x 6.02×10^23 molecules/mol = 1.88×10^17 molecules

Finally, we can calculate the surface area of the catalyst covered by these molecules:

A = N x a

where a is the area covered by a nitrogen molecule (0.162 nm^2), converted to m^2.

a = 0.162 nm^2 x (10^-18 m^2/nm^2) = 1.62×10^-20 m^2

A = 1.88×10^17 molecules x 1.62×10^-20 m^2/molecule = 3.05×10^-3 m^2

Therefore, the surface area of the catalyst covered by the nitrogen molecules is approximately 3.05×10^-3 m^2.

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