Question

Molar Mass (g/mol)
HNO3 63.013
N atom 14.007
H atom 1.0080
O atom 15.999
Avogadro's No. (mol-1)
6.022×1023

Answer 1

If a sample of HNO3 is found to contain 6.74 · 1024 N atoms how many moles of H atoms are present?

Answer 2

n=N/L = 6,74 *10^24 / 6,022*10^23 mol-1= 11,19 mol of "N" and ratio of N and H in HNO3 is same so n(H) = 11,19 mol