Find the number of hydrogen atoms present in $6.0 \mathrm{~g}$ of
Find the number of hydrogen atoms present in $6.0 \mathrm{~g}$ of

- $2.4 \times 10^{23}$
- $4.06 \times 10^{23}$
- $2.16 \times 10^{23}$
- $3.01 \times 10^{23}$
Solution
Moles of urea $=6.0 / 60=0.1$
Therefore moles of $\mathrm{H}$-atoms $=0.1 \times 4=0.4$ moles
Or $0.4 \times 6.02 \times 10^{23}=2.4 \times 10^{23}$
Asked in: MHT CET 2022 (11 Aug Shift 1)
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