Question

calculate the number of atoms of hydrogen present in 5.6 g of urea (NH2)2CO. Also calculate the number of atoms of N, C and O.​

Answer 1

Hello dear,

◆ Answer -

Number of H-atoms = 2.25×10^23 atoms

Number of N-atoms = 1.124×10^23 atoms

Number of C-atoms = 5.62×10^22 atoms

Number of O-atoms = 5.62×10^22 atoms

◆ Explaination -

Moles of urea are calculated as -

n = W / M

n = 5.6 / 60

n = 0.0933 mol

No of molecules of urea will be -

N(urea) = n × NA

N(urea) = 0.9333 × 6.022×10^23

N(urea) = 5.62×10^22 molecules

Each molecule of urea contains 4 H-atoms. Thus, no of H-atoms in 5.6 g urea are -

N(H) = 4 × N(urea)

N(H) = 4 × 5.62×10^22

N(H) = 2.25×10^23 atoms

Each molecule of urea contains 2 N-atoms. Thus, no of N-atoms in 5.6 g urea are -

N(N) = 2 × N(urea)

N(N) = 2 × 5.62×10^22

N(N) = 1.124×10^23 atoms

Each molecule of urea contains 1 C-atom. Thus, no of C-atoms in 5.6 g urea are -

N(C) = 1 × N(urea)

N(C) = 1 × 5.62×10^22

N(C) = 5.62×10^22 atoms

Each molecule of urea contains 1 O-atom. Thus, no of O-atoms in 5.6 g urea are -

N(O) = 1 × N(urea)

N(O) = 1 × 5.62×10^22

N(O) = 5.62×10^22 atoms

Thanks dear...