Question

Q. Calculate the number of atoms of
hydrogen present in 5.6 g of urea,
(NH,),CO. Also calculate the number
of atoms of N, C and O.
(Ans. : No. of atoms of H = 2.24 x 1023
N=1.124 x 1023 and C = 0.562 x 1023, O
= 0.562 x 1023)​

Answer 1

urea is CH4N2O

molar mass of urea = 12+1 *4+14*2+16

=60 gram

now moles of urea in 5.6 gram 7= given mass / molar mass

= 5.6 /60

=7/75

Now we know that moles = N/NA

=> 7/75=N/NA

=> N= 7NA/75 molecules

now 1 molecule of CH4N2O contain 4 hydrogen atom, 2 N atom, 1 O atom, and 1 C atom

• no of H atom in 7NA/75 molecule => 4 x 7NA/75 => 2.248 x 10^23
• similarly, no of N atom = 2 x 7 NA/75 = 1.124 x 10^23
• no. of C atom =1 x 7NA/75 =0.56 x 10^23
• no. of O atom 1 x 7NA/75=0.56x10^23