Question

24) Calculate the number of atoms of hydrogen present in 5.6 g of urea, (NHCO. Also calculate
the number of atoms of N, C and o.​

Answer 1

Answer:

Na= 6.022×10^23

43gm.......….. Na atoms

5.6 gm........Na×5.6/43 atoms of H

Explanation:

for N,C,O also same H.= Na×5.6/43 atoms

Answer 2

Explanation:

Molecular formula of urea isCH4N2O = molecular mass = 60g

Given mass ofCH4N2O = 5.6 g

= Moles of CH4N2O = 5.6/60 = 0.093.

Now for calculation of atoms.

For carbon atoms:

In 1 moleCH4N2O, moles of carbon = 1

= In 0.093 molesCH4N2O, moles of carbon = 0.093

= No of carbon atoms = 0.093*6.022*1023 = 5.60046 * 1022 atoms

For Hydrogen atoms:

In 1 moleCH4N2O, moles of hydrogen = 4

= In 0.093 moles ofCH4N2O, moles of hydrogen = 0.093 *4 = 0.372

= No of hydrogen atoms = 0.372*6.022*1023 = 2.240184 * 1023 atoms

For nitrogen atoms:

In 1 moleCH4N2O, moles of nitrogen = 2

= in 0.093 moles. moles of nitrogen = 0.093*2 = 0.186

= No of nitrogen atoms = 0.186*6.022*1023 = 1.120092 * 1023 atoms

For oxygen atoms:

In 1 moleCH4N2O, moles of oxygen = 1

= In 0.093 molesCH4N2O, moles of oxygen = 0.093

=No of oxygen atoms = 0.093*6.022*1023= 5.60046 * 1022atoms

Hope this helps!