Question

Calculate the total number of oxygen and nitrogen atoms present in a mixture containing 8.2g of calcium 10 millimoles of sodium nitrate Na/6 molecules of no2

Answer 1

Dear student ,

The number of moles of sodium nitrate = 10$\times$10^-3 = 10^-2 moles = 0.01 moles
Again the number of moles of NO₂ = $\frac{N_A/6}{N_A}$ = 0.167 moles
Total number of moles of N = 0.01 + 0.167 = 0.177 moles or 0.177$\times$6.02$\times$10²³ atoms = 1.0655$\times$10²³ number of atoms
Again the total number of moles of O = 3$\times$0.01 + 2$\times$0.167 = 0.364 moles or 0.364$\times$6.02$\times$10²³ atoms
= 2.19$\times$10²³ number of atoms
Regards