Calculate the number of atoms of hydrogen present in 5.6 g of urea. Also calculate the number of atoms of N, C and O.

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Solution

$U r e a \to C O (N H_{2})_{2}$
Given weight = $5.6 g$
$∴$ Moles of urea $= \frac{5.6}{60} = 0.0933$ moles
No. of $H$ atoms $= 4 \times 0.0933 \times N a = 2.25 \times 10^{23} a t o m s$
No. of $N$ atoms $= 2 \times 0.0933 \times N a = 1.123 \times 10^{23} a t o m s$
No. of $C$ atoms $= N o . o f O a t o m s = 0.0933 \times N a = 5.6 \times 10^{22} a t o m s$

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