What is the mass of sodium nitrite, which contains 1.8 * 10 ^ 24 oxygen atoms

Given:
N (O in NaNO2) = 1.8 * 10 ^ 24
To find:
m (NaNO2) -?
Decision:
1) Calculate the molar mass of NaNO2:
M (NaNO2) = Mr (NaNO2) = Ar (Na) * N (Na) + Ar (N) * N (N) + Ar (O) * N (O) = 23 * 1 + 14 * 1 + 16 * 2 = 69 g / mol;
2) Calculate the amount of substance O in NaNO2:
n (O in NaNO2) = N (O in NaNO2) / NA = 1.8 * 10 ^ 24 / 6.02 * 10 ^ 23 = 2.99 mol;
3) Calculate the amount of NaNO2 substance:
n (NaNO2) = n (O in NaNO2) / 2 = 2.99 / 2 = 1.495 mol;
4) Calculate the mass of NaNO2:
m (NaNO2) = n (NaNO2) * M (NaNO2) = 1.495 * 69 = 103.16 g.
Answer: The mass of NaNO2 is 103.16 g.