Question

Calculate number of atoms of hydrogen present in 5.6 grams of urea

Answer 1

Hiiii...

Urea is NH2- CO -NH2 ......( M.M = 60 )

No. of moles = Mass / Molar mass ...

Moles = 5.6 / 60 = 0.093 ....

No. of H atoms = N × Moles × 4 .... ( N is the avogadro number ....) ..

No. of H atoms = 0.093 × 4 × 6 × 10^23 ..
= 1.116 × 10^23 ......

I hope it helps uuuu......

Answer 2

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!