Question

calculate the mass of 1.0505*10^23 molecules of co2 the number of molecules in 0.25 moles of NH3​

Answer 1

MM= CO2. =44u

molecules= 1.0505×10^23

given mass= 44 × 6.022×10 23

1.0505×10 23

44×4= 176g

MM NH3= 14+1×3 = 17u

moles 1÷4

given mass = 17×0.25= 4.25g