Acommercial refrigeration unit accidentally releases (6.05x10^1) ml of ammonia gas at satp determine the mass and number of molecules of ammonia released

The volume given likely has an error, but I will work with it to show you how to solve this problem, so you will be able to find the right answer using the same procedure.

V = 6.05 * 10 ml = 0.00605 l

SATP => T = 373.15 K, P = 1 atm

pV = nRT => n = pV / RT

R = 0.0821 atm*l / (K*mol)

n = 1 atm * 0.00605 l / [ (0.0821 atm * l / K*mol) * 273.15 K] = 0.00026978 mol

So, the number of molecules is the number of moles times Avogadro's number:

0.0026978mol = 0.0026978 * 6.022*10^23 molecules = 1.6225 * 10^ 21 molecules

The mass is found using mass = molar mass * number of moles

Ammonia gas = NH3 => molar mass = 14 g/mol + 3 * 1g/mol = 17 g/mol

Mass = 17 g/mol * 0.0026978 mol = 0.4586 g.

With that procedure, you just have to replace the right volume of ammonia.