A commercial refrigeration unit accidentally releases (7.39*10^1) mL of ammonia gas at SATP. Determine the mass and number of molecules of ammonia gas released.

Question

Please show all the work, because I am getting 2 different answers.

Note: Your answer is assumed to be reduced to the highest power possible.

Answer 1

Here is one way to solve it:

T = 298.15K

P = 1 bar = 100.00 kPa

V = 7.39x101 mL = 73.9 mL = 0.0739 L NH₃

Rearrange the ideal gas law to for n PV = nRT

n = PV/RT = (100.00 kPa)(0.0739 L)/(8.314 L-kPa/Kmol)(298.15K)

n = 0.002981 moles NH₃

To calculate mass:

mass = 0.002981 moles NH₃ x 17.031 g/mol = 0.05077 g = 0.0508g (3 sig. figs.)

To calculate molecules:

number of molecules = 0.002981 moles x 6.02x10²³ molecules/mole = 1.79x10²¹ molecules

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Answer 2

SATP = standard ambient temperature and pressure = 298.15K and 1 bar (100.00 kPa)

7.39x101 ml = 73.9 ml = 0.0739 L NH₃

Using the ideal gas law: PV = nRT and solving for n (moles) we have...

n = PV/RT = (100.00 kPa)(0.0739 L)/(8.314 L-kPa/Kmol)(298.15K)

n = 0.002981 moles NH₃

mass = 0.002981 moles NH₃ x 17.031 g/mol = 0.05077 g = 0.0508g (3 sig. figs.)

number of molecules = 0.002981 moles x 6.02x10²³ molecules/mole = 1.79x10²¹ molecules